Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games

Opening subject page...

Loading your content

  1. My Subjects
  3. AP Chemistry

  5. Flashcards

AP Chemistry Flashcards: Strong Acids And Bases Ph Poh

Study Strong Acids And Bases Ph Poh in AP Chemistry with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

← Back to flashcard decks

What this deck covers

This deck focuses on Strong Acids And Bases Ph Poh, 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.

AP Chemistry Flashcards: Strong Acids And Bases Ph Poh

1

/ 30

0 reviewed

0% Complete

0 reviewing
QUESTION

What is KwK_wKw​ at 25∘C25^\circ\text{C}25∘C for water?

Tap or drag to reveal answer

ANSWER

Kw=[H+][OH−]=1.0×10−14K_w=[\text{H}^+][\text{OH}^-]=1.0\times10^{-14}Kw​=[H+][OH−]=1.0×10−14. Water's ion product constant at 25°C.

Swipe Right = I Know It! 🎉

Swipe Left = Still Learning

All flashcards

Flashcard 1: What is KwK_wKw​ at 25∘C25^\circ\text{C}25∘C for water?

Answer: Kw=[H+][OH−]=1.0×10−14K_w=[\text{H}^+][\text{OH}^-]=1.0\times10^{-14}Kw​=[H+][OH−]=1.0×10−14. Water's ion product constant at 25°C.

Flashcard 2: What is the [OH−][\text{OH}^-][OH−] in a solution with [H+]=1.0×10−4 M[\text{H}^+]=1.0\times10^{-4}\,\text{M}[H+]=1.0×10−4M at 25∘C25^\circ\text{C}25∘C?

Answer: [OH−]=1.0×10−10 M[\text{OH}^-]=1.0\times10^{-10}\,\text{M}[OH−]=1.0×10−10M. Use Kw=[H⁺][OH⁻]=1.0×10⁻¹⁴; [OH⁻]=1.0×10⁻¹⁴/1.0×10⁻⁴.

Flashcard 3: What is the pH\text{pH}pH of a 0.10 M0.10\,\text{M}0.10M strong base that provides 3 OH−3\,\text{OH}^-3OH− per formula unit?

Answer: pH=13.48\text{pH}=13.48pH=13.48. n=3, so [OH⁻]=3×0.10=0.30 M; pOH=0.52; pH=14-0.52=13.48.

Flashcard 4: What does nnn represent in [OH−]≈n Cbase[\text{OH}^-]\approx n\,C_{\text{base}}[OH−]≈nCbase​ for a strong base?

Answer: n=moles of OH− produced per mole of basen=\text{moles of }\text{OH}^-\text{ produced per mole of base}n=moles of OH− produced per mole of base. Counts OH⁻ ions released per base molecule (e.g., 2 for Ba(OH)₂).

Flashcard 5: What assumption is used for a strong base to relate [OH−][\text{OH}^-][OH−] to molarity?

Answer: [OH−]≈n Cbase[\text{OH}^-]\approx n\,C_{\text{base}}[OH−]≈nCbase​. Strong bases fully dissociate; n is OH⁻ ions per base molecule.

Flashcard 6: What is the [OH−][\text{OH}^-][OH−] when pOH=9.00\text{pOH}=9.00pOH=9.00?

Answer: [OH−]=1.0×10−9 M[\text{OH}^-]=1.0\times10^{-9}\,\text{M}[OH−]=1.0×10−9M. Apply [OH⁻]=10⁻ᵖᴼᴴ=10⁻⁹=1.0×10⁻⁹ M.

Flashcard 7: What is the formula that relates pH\text{pH}pH to [H+][\text{H}^+][H+]?

Answer: pH=−log⁡[H+]\text{pH}=-\log[\text{H}^+]pH=−log[H+]. Negative log converts hydrogen ion concentration to pH scale.

Flashcard 8: What is the relationship between pH\text{pH}pH and pOH\text{pOH}pOH at 25∘C25^\circ\text{C}25∘C?

Answer: pH+pOH=14.00\text{pH}+\text{pOH}=14.00pH+pOH=14.00. At 25°C, the sum always equals 14 due to water's ion product.

Flashcard 9: What is the pOH\text{pOH}pOH when pH=3.25\text{pH}=3.25pH=3.25 at 25∘C25^\circ\text{C}25∘C?

Answer: pOH=10.75\text{pOH}=10.75pOH=10.75. Use pH+pOH=14.00; pOH=14.00-3.25=10.75.

Flashcard 10: What is the pH\text{pH}pH when pOH=1.60\text{pOH}=1.60pOH=1.60 at 25∘C25^\circ\text{C}25∘C?

Answer: pH=12.40\text{pH}=12.40pH=12.40. Use pH+pOH=14.00; pH=14.00-1.60=12.40.

Flashcard 11: What is the formula that relates pOH\text{pOH}pOH to [OH−][\text{OH}^-][OH−]?

Answer: pOH=−log⁡[OH−]\text{pOH}=-\log[\text{OH}^-]pOH=−log[OH−]. Negative log converts hydroxide ion concentration to pOH scale.

Flashcard 12: What is the pOH of a solution with [H3O+]=1.0×10−9 M[\mathrm{H_3O^+}]=1.0\times10^{-9}\ \mathrm{M}[H3​O+]=1.0×10−9 M at 25∘C25^\circ\mathrm{C}25∘C?

Answer: pOH=5.00\mathrm{pOH}=5.00pOH=5.00. pH=−log⁡(1.0×10−9)=9\mathrm{pH}=-\log(1.0\times10^{-9})=9pH=−log(1.0×10−9)=9; pOH=14−9=5\mathrm{pOH}=14-9=5pOH=14−9=5.

Flashcard 13: What is [H3O+][\mathrm{H_3O^+}][H3​O+] for a solution with pH=2.50\mathrm{pH}=2.50pH=2.50?

Answer: [H3O+]=10−2.50=3.2×10−3 M[\mathrm{H_3O^+}]=10^{-2.50}=3.2\times10^{-3}\ \mathrm{M}[H3​O+]=10−2.50=3.2×10−3 M. Use [H3O+]=10−pH=10−2.50[\mathrm{H_3O^+}]=10^{-\mathrm{pH}}=10^{-2.50}[H3​O+]=10−pH=10−2.50.

Flashcard 14: What is the pH of a 0.015 M0.015\ \mathrm{M}0.015 M solution of Ba(OH)2\mathrm{Ba(OH)_2}Ba(OH)2​ at 25∘C25^\circ\mathrm{C}25∘C?

Answer: pH=12.48\mathrm{pH}=12.48pH=12.48. pH=14.00−pOH=14.00−1.52=12.48\mathrm{pH}=14.00-\mathrm{pOH}=14.00-1.52=12.48pH=14.00−pOH=14.00−1.52=12.48.

Flashcard 15: What is the pOH of a 0.015 M0.015\ \mathrm{M}0.015 M solution of Ba(OH)2\mathrm{Ba(OH)_2}Ba(OH)2​?

Answer: pOH=1.52\mathrm{pOH}=1.52pOH=1.52. pOH=−log⁡(0.030)=−log⁡(3×10−2)=1.52\mathrm{pOH}=-\log(0.030)=-\log(3\times10^{-2})=1.52pOH=−log(0.030)=−log(3×10−2)=1.52.

Flashcard 16: What is the pH of a 2.0×10−5 M2.0\times10^{-5}\ \mathrm{M}2.0×10−5 M solution of HBr\mathrm{HBr}HBr?

Answer: pH=4.70\mathrm{pH}=4.70pH=4.70. pH=−log⁡(2.0×10−5)=5−log⁡(2)=4.70\mathrm{pH}=-\log(2.0\times10^{-5})=5-\log(2)=4.70pH=−log(2.0×10−5)=5−log(2)=4.70.

Flashcard 17: What is [OH−][\mathrm{OH^-}][OH−] for a 0.015 M0.015\ \mathrm{M}0.015 M solution of Ba(OH)2\mathrm{Ba(OH)_2}Ba(OH)2​?

Answer: [OH−]=0.030 M[\mathrm{OH^-}]=0.030\ \mathrm{M}[OH−]=0.030 M. Ba(OH)₂ releases 2 OH⁻ per unit: 0.015×2=0.030 M0.015\times^2=0.030\ \mathrm{M}0.015×2=0.030 M.

Flashcard 18: What is the pH of a 4.0×10−2 M4.0\times10^{-2}\ \mathrm{M}4.0×10−2 M solution of NaOH\mathrm{NaOH}NaOH at 25∘C25^\circ\mathrm{C}25∘C?

Answer: pH=12.60\mathrm{pH}=12.60pH=12.60. pH=14.00−pOH=14.00−1.40=12.60\mathrm{pH}=14.00-\mathrm{pOH}=14.00-1.40=12.60pH=14.00−pOH=14.00−1.40=12.60.

Flashcard 19: What is the pOH of a 4.0×10−2 M4.0\times10^{-2}\ \mathrm{M}4.0×10−2 M solution of NaOH\mathrm{NaOH}NaOH?

Answer: pOH=1.40\mathrm{pOH}=1.40pOH=1.40. pOH=−log⁡(4.0×10−2)=2−log⁡(4)=1.40\mathrm{pOH}=-\log(4.0\times10^{-2})=2-\log(4)=1.40pOH=−log(4.0×10−2)=2−log(4)=1.40.

Flashcard 20: Identify the pH of pure water at 25∘C25^\circ\mathrm{C}25∘C.

Answer: pH=7.00\mathrm{pH}=7.00pH=7.00. Neutral water has [H3O+]=[OH−]=1.0×10−7 M[\mathrm{H_3O^+}]=[\mathrm{OH^-}]=1.0\times10^{-7}\ \mathrm{M}[H3​O+]=[OH−]=1.0×10−7 M.

Flashcard 21: What is the pH of a 3.0×10−4 M3.0\times10^{-4}\ \mathrm{M}3.0×10−4 M solution of Ca(OH)2\mathrm{Ca(OH)_2}Ca(OH)2​ at 25∘C25^\circ\mathrm{C}25∘C?

Answer: pH=10.78\mathrm{pH}=10.78pH=10.78. [OH−]=2×3.0×10−4=6.0×10−4[\mathrm{OH^-}]=2\times^3.0\times10^{-4}=6.0\times10^{-4}[OH−]=2×3.0×10−4=6.0×10−4; pOH=3.22\mathrm{pOH}=3.22pOH=3.22; pH=10.78\mathrm{pH}=10.78pH=10.78.

Flashcard 22: State the expression for KwK_wKw​ and its value at 25∘C25^\circ\mathrm{C}25∘C.

Answer: Kw=[H3O+][OH−]=1.0×10−14K_w=[\mathrm{H_3O^+}][\mathrm{OH^-}]=1.0\times10^{-14}Kw​=[H3​O+][OH−]=1.0×10−14. Water autoionization constant at 25°C25°\mathrm{C}25°C.

Flashcard 23: Identify the assumption used for strong acids when finding [H3O+][\mathrm{H_3O^+}][H3​O+].

Answer: Strong acids dissociate ≈100%\approx100\%≈100%: [H3O+]≈[\mathrm{H_3O^+}]\approx[H3​O+]≈ acid molarity. Complete dissociation means initial acid concentration equals [H3O+][\mathrm{H_3O^+}][H3​O+].

Flashcard 24: Identify the assumption used for strong bases when finding [OH−][\mathrm{OH^-}][OH−].

Answer: Strong bases dissociate ≈100%\approx100\%≈100%: [OH−]≈[\mathrm{OH^-}]\approx[OH−]≈ base molarity ×\times× stoichiometry. Complete dissociation; multiply by OH⁻ per formula unit.

Flashcard 25: What is [H3O+][\mathrm{H_3O^+}][H3​O+] for a 0.030 M0.030\ \mathrm{M}0.030 M solution of HCl\mathrm{HCl}HCl?

Answer: [H3O+]=0.030 M[\mathrm{H_3O^+}]=0.030\ \mathrm{M}[H3​O+]=0.030 M. HCl is monoprotic strong acid, so [H3O+][\mathrm{H_3O^+}][H3​O+] equals molarity.

Flashcard 26: What is the pH of a 1.0×10−3 M1.0\times10^{-3}\ \mathrm{M}1.0×10−3 M solution of HNO3\mathrm{HNO_3}HNO3​?

Answer: pH=3.00\mathrm{pH}=3.00pH=3.00. pH=−log⁡(1.0×10−3)=3.00\mathrm{pH}=-\log(1.0\times10^{-3})=3.00pH=−log(1.0×10−3)=3.00 for strong acid HNO₃.

Flashcard 27: State the formula for pH in terms of hydronium concentration.

Answer: pH=−log⁡[H3O+]\mathrm{pH}=-\log[\mathrm{H_3O^+}]pH=−log[H3​O+]. Negative log converts small [H3O+][\mathrm{H_3O^+}][H3​O+] values to manageable pH scale.

Flashcard 28: State the formula for pOH in terms of hydroxide concentration.

Answer: pOH=−log⁡[OH−]\mathrm{pOH}=-\log[\mathrm{OH^-}]pOH=−log[OH−]. Negative log converts small [OH−][\mathrm{OH^-}][OH−] values to manageable pOH scale.

Flashcard 29: What is the relationship between pH and pOH at 25∘C25^\circ\mathrm{C}25∘C?

Answer: pH+pOH=14.00\mathrm{pH}+\mathrm{pOH}=14.00pH+pOH=14.00. At 25°C25°\mathrm{C}25°C, this sum equals −log⁡(Kw)=14.00-\log(K_w)=14.00−log(Kw​)=14.00.

Flashcard 30: What is the pH of a solution with [OH−]=1.0×10−12 M[\mathrm{OH^-}]=1.0\times10^{-12}\ \mathrm{M}[OH−]=1.0×10−12 M at 25∘C25^\circ\mathrm{C}25∘C?

Answer: pH=2.00\mathrm{pH}=2.00pH=2.00. pOH=−log⁡(1.0×10−12)=12\mathrm{pOH}=-\log(1.0\times10^{-12})=12pOH=−log(1.0×10−12)=12; pH=14−12=2\mathrm{pH}=14-12=2pH=14−12=2.