AP Chemistry Flashcards: Ph And Solubility

What this deck covers

This deck focuses on Ph And Solubility, giving you a quick way to review the definitions, rules, and examples that matter most for AP Chemistry.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Identify the pH of a neutral solution at 25°C.

Answer: pH = 7. Equal concentrations of $[H^+]$ and $[OH^-]$.

Flashcard 2: Determine the pOH of a solution with $[OH^-] = 1 \times 10^{-4} \text{ M}$.

Answer: pOH = 4. $pOH = -\log(10^{-4}) = 4$

Flashcard 3: What is the solubility product constant for $PbI_2$?

Answer: $K_{sp} = [Pb^{2+}][I^-]^2$. Iodide ion squared due to formula stoichiometry.

Flashcard 4: What is the pH of a $0.1 \text{ M}$ $NaOH$ solution?

Answer: pH = 13. Strong base: $[OH^-] = 0.1$, $pOH = 1$

Flashcard 5: Identify the pH of a neutral solution at 25°C.

Answer: pH = 7. Equal concentrations of $[H^+]$ and $[OH^-]$.

Flashcard 6: State the $K_{sp}$ expression for $ZnS$.

Answer: $K_{sp} = [Zn^{2+}][S^{2-}]$. 1:1 stoichiometry for ion products.

Flashcard 7: What is the solubility product constant for $PbI_2$?

Answer: $K_{sp} = [Pb^{2+}][I^-]^2$. Iodide ion squared due to formula stoichiometry.

Flashcard 8: What is the definition of the solubility product constant ($K_{sp}$)?

Answer: $K_{sp}$ is the equilibrium constant for a solid dissolving in water. Represents dissolution equilibrium of ionic solids.

Flashcard 9: What is the definition of the solubility product constant ($K_{sp}$)?

Answer: $K_{sp}$ is the equilibrium constant for a solid dissolving in water. Represents dissolution equilibrium of ionic solids.

Flashcard 10: State the formula for calculating pOH from $[OH^-]$.

Answer: $pOH = -\text{log}[OH^-]$. Take negative log of hydroxide ion concentration.

Flashcard 11: Determine the pH of a $0.025 \text{ M}$ $HBr$ solution.

Answer: pH = 1.6. $pH = -\log(0.025) = 1.6$

Flashcard 12: Which solution has a higher pH: $0.1 \text{ M}$ $HCl$ or $0.1 \text{ M}$ $NaOH$?

Answer: $0.1 \text{ M}$ $NaOH$. $NaOH$ is basic with higher pH value.

Flashcard 13: Find the pH of a $0.075 \text{ M}$ $Ba(OH)_2$ solution.

Answer: pH = 13.18. Diprotic base: $[OH^-] = 0.15$, $pOH = 0.82$

Flashcard 14: What is the $K_{sp}$ expression for $CaF_2$?

Answer: $K_{sp} = [Ca^{2+}][F^-]^2$. Fluoride ion squared due to formula stoichiometry.

Flashcard 15: State the formula for converting pOH to pH.

Answer: $pH = 14 - pOH$. Derived from $pH + pOH = 14$ relationship.

Flashcard 16: What is the $K_{sp}$ expression for $CuS$?

Answer: $K_{sp} = [Cu^{2+}][S^{2-}]$. 1:1 stoichiometry for ion products.

Flashcard 17: What is the pH of a $0.02 \text{ M}$ $KOH$ solution?

Answer: pH = 12.3. $[OH^-] = 0.02$, $pOH = 1.7$

Flashcard 18: What is the $K_{sp}$ expression for $PbSO_4$?

Answer: $K_{sp} = [Pb^{2+}][SO_4^{2-}]$. 1:1 stoichiometry for ion products.

Flashcard 19: What is the $K_{sp}$ expression for $Mg(OH)_2$?

Answer: $K_{sp} = [Mg^{2+}][OH^-]^2$. Hydroxide ion squared due to formula stoichiometry.

Flashcard 20: What is the pH of a solution with $[H^+] = 1 \times 10^{-3} \text{ M}$?

Answer: pH = 3. $pH = -\log(10^{-3}) = 3$

Flashcard 21: State the ion product constant for water ($K_w$) at 25°C.

Answer: $K_w = 1.0 \times 10^{-14}$. Equilibrium constant for water autoionization.

Flashcard 22: Identify the $K_{sp}$ expression for $BaSO_4$.

Answer: $K_{sp} = [Ba^{2+}][SO_4^{2-}]$. 1:1 stoichiometry for ion products.

Flashcard 23: Identify the $K_{sp}$ expression for $BaSO_4$.

Answer: $K_{sp} = [Ba^{2+}][SO_4^{2-}]$. 1:1 stoichiometry for ion products.

Flashcard 24: What is the pH of a $0.02 \text{ M}$ $KOH$ solution?

Answer: pH = 12.3. $[OH^-] = 0.02$, $pOH = 1.7$

Flashcard 25: What is the $K_{sp}$ expression for $CaCO_3$?

Answer: $K_{sp} = [Ca^{2+}][CO_3^{2-}]$. 1:1 stoichiometry for ion products.

Flashcard 26: What is the $K_{sp}$ expression for $Ag_2CrO_4$?

Answer: $K_{sp} = [Ag^+]^2[CrO_4^{2-}]$. Silver ion squared due to formula stoichiometry.

Flashcard 27: What is the relationship between pH and pOH at 25°C?

Answer: $pH + pOH = 14$. Sum equals 14 at standard temperature.

Flashcard 28: What is the definition of pH?

Answer: pH is the negative logarithm of the hydrogen ion concentration: $pH = -\text{log}[H^+]$. It measures acidity using logarithmic scale.

Flashcard 29: State the formula for calculating pH from $[H^+]$.

Answer: $pH = -\text{log}[H^+]$. Take negative log of hydrogen ion concentration.

Flashcard 30: What is the relationship between pH and pOH at 25°C?

Answer: $pH + pOH = 14$. Sum equals 14 at standard temperature.

Flashcard 31: What is the solubility product constant ($K_{sp}$) expression for $AgCl$?

Answer: $K_{sp} = [Ag^+][Cl^-]$. Products of ion concentrations at equilibrium.

Flashcard 32: What is the $K_{sp}$ expression for $Hg_2I_2$?

Answer: $K_{sp} = [Hg_2^{2+}][I^-]^2$. Iodide ion squared due to formula stoichiometry.

Flashcard 33: What is the solubility product constant for $PbI_2$?

Answer: $K_{sp} = [Pb^{2+}][I^-]^2$. Iodide ion squared due to formula stoichiometry.

Flashcard 34: What is the $K_{sp}$ expression for $Mg(OH)_2$?

Answer: $K_{sp} = [Mg^{2+}][OH^-]^2$. Hydroxide ion squared due to formula stoichiometry.

Flashcard 35: Calculate the pH of a $0.1 \text{ M}$ $CH_3COOH$ solution, $K_a = 1.8 \times 10^{-5}$.

Answer: pH ≈ 2.87. Weak acid requires ICE table calculation.

Flashcard 36: What is the definition of pH?

Answer: pH is the negative logarithm of the hydrogen ion concentration: $pH = -\text{log}[H^+]$. It measures acidity using logarithmic scale.

Flashcard 37: Determine the pOH of a solution with $[OH^-] = 1 \times 10^{-4} \text{ M}$.

Answer: pOH = 4. $pOH = -\log(10^{-4}) = 4$

Flashcard 38: What is the $K_{sp}$ expression for $CuS$?

Answer: $K_{sp} = [Cu^{2+}][S^{2-}]$. 1:1 stoichiometry for ion products.

Flashcard 39: Find the pH of a $0.03 \text{ M}$ $LiOH$ solution.

Answer: pH = 12.5. $[OH^-] = 0.03$, $pOH = 1.52$

Flashcard 40: Determine the pH of a $0.025 \text{ M}$ $HBr$ solution.

Answer: pH = 1.6. $pH = -\log(0.025) = 1.6$

Flashcard 41: What is the $K_{sp}$ expression for $Ag_2CrO_4$?

Answer: $K_{sp} = [Ag^+]^2[CrO_4^{2-}]$. Silver ion squared due to formula stoichiometry.

Flashcard 42: Determine the pOH of a solution with $[OH^-] = 1 \times 10^{-4} \text{ M}$.

Answer: pOH = 4. $pOH = -\log(10^{-4}) = 4$

Flashcard 43: What is the $K_{sp}$ expression for $Ca(OH)_2$?

Answer: $K_{sp} = [Ca^{2+}][OH^-]^2$. Hydroxide ion squared due to formula stoichiometry.

Flashcard 44: What is the $K_{sp}$ expression for $Hg_2I_2$?

Answer: $K_{sp} = [Hg_2^{2+}][I^-]^2$. Iodide ion squared due to formula stoichiometry.

Flashcard 45: Which solution has a higher pH: $0.1 \text{ M}$ $HCl$ or $0.1 \text{ M}$ $NaOH$?

Answer: $0.1 \text{ M}$ $NaOH$. $NaOH$ is basic with higher pH value.

Flashcard 46: Determine the $K_{sp}$ expression for $Fe(OH)_3$.

Answer: $K_{sp} = [Fe^{3+}][OH^-]^3$. Hydroxide ion cubed due to formula stoichiometry.

Flashcard 47: What is the $K_{sp}$ expression for $Ca(OH)_2$?

Answer: $K_{sp} = [Ca^{2+}][OH^-]^2$. Hydroxide ion squared due to formula stoichiometry.

Flashcard 48: What is the pH of a $0.01 \text{ M}$ $HCl$ solution?

Answer: pH = 2. Strong acid completely ionizes: $[H^+] = 0.01$

Flashcard 49: Determine the $K_{sp}$ expression for $Fe(OH)_3$.

Answer: $K_{sp} = [Fe^{3+}][OH^-]^3$. Hydroxide ion cubed due to formula stoichiometry.

Flashcard 50: Find the pH of a $0.03 \text{ M}$ $LiOH$ solution.

Answer: pH = 12.5. $[OH^-] = 0.03$, $pOH = 1.52$

Flashcard 51: State the formula for calculating pOH from $[OH^-]$.

Answer: $pOH = -\text{log}[OH^-]$. Take negative log of hydroxide ion concentration.

Flashcard 52: State the formula for converting pOH to pH.

Answer: $pH = 14 - pOH$. Derived from $pH + pOH = 14$ relationship.

Flashcard 53: Calculate the pH of a $0.005 \text{ M}$ $HCl$ solution.

Answer: pH = 2.3. $pH = -\log(0.005) = 2.3$

Flashcard 54: What is the definition of pH?

Answer: pH is the negative logarithm of the hydrogen ion concentration: $pH = -\text{log}[H^+]$. It measures acidity using logarithmic scale.

Flashcard 55: What is the pH of a $0.1 \text{ M}$ $NaOH$ solution?

Answer: pH = 13. Strong base: $[OH^-] = 0.1$, $pOH = 1$

Flashcard 56: What is the $K_{sp}$ expression for $Ca(OH)_2$?

Answer: $K_{sp} = [Ca^{2+}][OH^-]^2$. Hydroxide ion squared due to formula stoichiometry.

Flashcard 57: Determine the $K_{sp}$ expression for $Fe(OH)_3$.

Answer: $K_{sp} = [Fe^{3+}][OH^-]^3$. Hydroxide ion cubed due to formula stoichiometry.

Flashcard 58: Calculate the pH of pure water at 25°C.

Answer: pH = 7. $[H^+] = 1.0 \times 10^{-7}$ at equilibrium.

Flashcard 59: What is the pH of a $0.02 \text{ M}$ $KOH$ solution?

Answer: pH = 12.3. $[OH^-] = 0.02$, $pOH = 1.7$

Flashcard 60: Calculate the pH of a $0.001 \text{ M}$ $HNO_3$ solution.

Answer: pH = 3. Strong acid: $pH = -\log(0.001) = 3$

Flashcard 61: What is the $K_{sp}$ expression for $Ag_2SO_4$?

Answer: $K_{sp} = [Ag^+]^2[SO_4^{2-}]$. Silver ion squared due to formula stoichiometry.

Flashcard 62: What is the $K_{sp}$ expression for $PbSO_4$?

Answer: $K_{sp} = [Pb^{2+}][SO_4^{2-}]$. 1:1 stoichiometry for ion products.

Flashcard 63: State the formula for calculating pOH from $[OH^-]$.

Answer: $pOH = -\text{log}[OH^-]$. Take negative log of hydroxide ion concentration.

Flashcard 64: Which solution has a higher pH: $0.1 \text{ M}$ $HCl$ or $0.1 \text{ M}$ $NaOH$?

Answer: $0.1 \text{ M}$ $NaOH$. $NaOH$ is basic with higher pH value.

Flashcard 65: State the ion product constant for water ($K_w$) at 25°C.

Answer: $K_w = 1.0 \times 10^{-14}$. Equilibrium constant for water autoionization.

Flashcard 66: Find the pH of a $0.03 \text{ M}$ $LiOH$ solution.

Answer: pH = 12.5. $[OH^-] = 0.03$, $pOH = 1.52$

Flashcard 67: What is the pH of a $0.1 \text{ M}$ $NaOH$ solution?

Answer: pH = 13. Strong base: $[OH^-] = 0.1$, $pOH = 1$

Flashcard 68: Which is more soluble in water: $AgCl$ or $NaCl$?

Answer: $NaCl$. $NaCl$ is highly soluble ionic compound.

Flashcard 69: What is the $K_{sp}$ expression for $Ag_2CrO_4$?

Answer: $K_{sp} = [Ag^+]^2[CrO_4^{2-}]$. Silver ion squared due to formula stoichiometry.

Flashcard 70: Calculate the pH of a $0.1 \text{ M}$ $CH_3COOH$ solution, $K_a = 1.8 \times 10^{-5}$.

Answer: pH ≈ 2.87. Weak acid requires ICE table calculation.

Flashcard 71: State the formula for calculating pH from $[H^+]$.

Answer: $pH = -\text{log}[H^+]$. Take negative log of hydrogen ion concentration.

Flashcard 72: Calculate the pH of a $0.001 \text{ M}$ $HNO_3$ solution.

Answer: pH = 3. Strong acid: $pH = -\log(0.001) = 3$

Flashcard 73: What is the solubility product constant ($K_{sp}$) expression for $AgCl$?

Answer: $K_{sp} = [Ag^+][Cl^-]$. Products of ion concentrations at equilibrium.

Flashcard 74: What is the $K_{sp}$ expression for $CaF_2$?

Answer: $K_{sp} = [Ca^{2+}][F^-]^2$. Fluoride ion squared due to formula stoichiometry.

Flashcard 75: Identify the $K_{sp}$ expression for $BaSO_4$.

Answer: $K_{sp} = [Ba^{2+}][SO_4^{2-}]$. 1:1 stoichiometry for ion products.

Flashcard 76: Calculate the pH of pure water at 25°C.

Answer: pH = 7. $[H^+] = 1.0 \times 10^{-7}$ at equilibrium.

Flashcard 77: Find the pH of a $0.075 \text{ M}$ $Ba(OH)_2$ solution.

Answer: pH = 13.18. Diprotic base: $[OH^-] = 0.15$, $pOH = 0.82$

Flashcard 78: What is the pH of a solution with $[H^+] = 1 \times 10^{-7} \text{ M}$?

Answer: pH = 7. $pH = -\log(10^{-7}) = 7$

Flashcard 79: What is the pH of a solution with $[H^+] = 1 \times 10^{-3} \text{ M}$?

Answer: pH = 3. $pH = -\log(10^{-3}) = 3$

Flashcard 80: What is the pH of a $0.01 \text{ M}$ $HCl$ solution?

Answer: pH = 2. Strong acid completely ionizes: $[H^+] = 0.01$

Flashcard 81: What is the $K_{sp}$ expression for $CuS$?

Answer: $K_{sp} = [Cu^{2+}][S^{2-}]$. 1:1 stoichiometry for ion products.

Flashcard 82: What is the pH of a $0.01 \text{ M}$ $HCl$ solution?

Answer: pH = 2. Strong acid completely ionizes: $[H^+] = 0.01$

Flashcard 83: State the formula for converting pOH to pH.

Answer: $pH = 14 - pOH$. Derived from $pH + pOH = 14$ relationship.

Flashcard 84: What is the $K_{sp}$ expression for $Mg(OH)_2$?

Answer: $K_{sp} = [Mg^{2+}][OH^-]^2$. Hydroxide ion squared due to formula stoichiometry.

Flashcard 85: What is the pH of a solution with $[H^+] = 1 \times 10^{-7} \text{ M}$?

Answer: pH = 7. $pH = -\log(10^{-7}) = 7$

Flashcard 86: Determine the pH of a $0.025 \text{ M}$ $HBr$ solution.

Answer: pH = 1.6. $pH = -\log(0.025) = 1.6$

Flashcard 87: What is the pH of a solution with $[H^+] = 1 \times 10^{-3} \text{ M}$?

Answer: pH = 3. $pH = -\log(10^{-3}) = 3$

Flashcard 88: What is the $K_{sp}$ expression for $Ag_2SO_4$?

Answer: $K_{sp} = [Ag^+]^2[SO_4^{2-}]$. Silver ion squared due to formula stoichiometry.

Flashcard 89: Identify the pH of a neutral solution at 25°C.

Answer: pH = 7. Equal concentrations of $[H^+]$ and $[OH^-]$.

Flashcard 90: If $[OH^-] = 1 \times 10^{-9} \text{ M}$, what is the pH of the solution?

Answer: pH = 5. $pOH = 9$, so $pH = 14 - 9 = 5$

Flashcard 91: Calculate the pH of a $0.1 \text{ M}$ $CH_3COOH$ solution, $K_a = 1.8 \times 10^{-5}$.

Answer: pH ≈ 2.87. Weak acid requires ICE table calculation.

Flashcard 92: What is the $K_{sp}$ expression for $Hg_2I_2$?

Answer: $K_{sp} = [Hg_2^{2+}][I^-]^2$. Iodide ion squared due to formula stoichiometry.

Flashcard 93: What is the $K_{sp}$ expression for $CaF_2$?

Answer: $K_{sp} = [Ca^{2+}][F^-]^2$. Fluoride ion squared due to formula stoichiometry.

Flashcard 94: What is the solubility product constant ($K_{sp}$) expression for $AgCl$?

Answer: $K_{sp} = [Ag^+][Cl^-]$. Products of ion concentrations at equilibrium.

Flashcard 95: State the formula for calculating pH from $[H^+]$.

Answer: $pH = -\text{log}[H^+]$. Take negative log of hydrogen ion concentration.

Flashcard 96: Calculate the pH of pure water at 25°C.

Answer: pH = 7. $[H^+] = 1.0 \times 10^{-7}$ at equilibrium.

Flashcard 97: If $[OH^-] = 1 \times 10^{-9} \text{ M}$, what is the pH of the solution?

Answer: pH = 5. $pOH = 9$, so $pH = 14 - 9 = 5$

Flashcard 98: Find the pH of a $0.075 \text{ M}$ $Ba(OH)_2$ solution.

Answer: pH = 13.18. Diprotic base: $[OH^-] = 0.15$, $pOH = 0.82$

Flashcard 99: Calculate the pH of a $0.001 \text{ M}$ $HNO_3$ solution.

Answer: pH = 3. Strong acid: $pH = -\log(0.001) = 3$

Flashcard 100: If $[OH^-] = 1 \times 10^{-9} \text{ M}$, what is the pH of the solution?

Answer: pH = 5. $pOH = 9$, so $pH = 14 - 9 = 5$