Strong Acids and Bases, pH, pOH - AP Chemistry
Card 1 of 30
What is $K_w$ at $25^\circ\text{C}$ for water?
What is $K_w$ at $25^\circ\text{C}$ for water?
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$K_w=[\text{H}^+][\text{OH}^-]=1.0\times10^{-14}$. Water's ion product constant at 25°C.
$K_w=[\text{H}^+][\text{OH}^-]=1.0\times10^{-14}$. Water's ion product constant at 25°C.
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What is the $[\text{OH}^-]$ in a solution with $[\text{H}^+]=1.0\times10^{-4},\text{M}$ at $25^\circ\text{C}$?
What is the $[\text{OH}^-]$ in a solution with $[\text{H}^+]=1.0\times10^{-4},\text{M}$ at $25^\circ\text{C}$?
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$[\text{OH}^-]=1.0\times10^{-10},\text{M}$. Use Kw=[H⁺][OH⁻]=1.0×10⁻¹⁴; [OH⁻]=1.0×10⁻¹⁴/1.0×10⁻⁴.
$[\text{OH}^-]=1.0\times10^{-10},\text{M}$. Use Kw=[H⁺][OH⁻]=1.0×10⁻¹⁴; [OH⁻]=1.0×10⁻¹⁴/1.0×10⁻⁴.
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What is the $\text{pH}$ of a $0.10,\text{M}$ strong base that provides $3,\text{OH}^-$ per formula unit?
What is the $\text{pH}$ of a $0.10,\text{M}$ strong base that provides $3,\text{OH}^-$ per formula unit?
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$\text{pH}=13.48$. n=3, so [OH⁻]=3×0.10=0.30 M; pOH=0.52; pH=14-0.52=13.48.
$\text{pH}=13.48$. n=3, so [OH⁻]=3×0.10=0.30 M; pOH=0.52; pH=14-0.52=13.48.
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What does $n$ represent in $[\text{OH}^-]\approx n,C_{\text{base}}$ for a strong base?
What does $n$ represent in $[\text{OH}^-]\approx n,C_{\text{base}}$ for a strong base?
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$n=\text{moles of }\text{OH}^-\text{ produced per mole of base}$. Counts OH⁻ ions released per base molecule (e.g., 2 for Ba(OH)₂).
$n=\text{moles of }\text{OH}^-\text{ produced per mole of base}$. Counts OH⁻ ions released per base molecule (e.g., 2 for Ba(OH)₂).
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What assumption is used for a strong base to relate $[\text{OH}^-]$ to molarity?
What assumption is used for a strong base to relate $[\text{OH}^-]$ to molarity?
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$[\text{OH}^-]\approx n,C_{\text{base}}$. Strong bases fully dissociate; n is OH⁻ ions per base molecule.
$[\text{OH}^-]\approx n,C_{\text{base}}$. Strong bases fully dissociate; n is OH⁻ ions per base molecule.
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What is the $[\text{OH}^-]$ when $\text{pOH}=9.00$?
What is the $[\text{OH}^-]$ when $\text{pOH}=9.00$?
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$[\text{OH}^-]=1.0\times10^{-9},\text{M}$. Apply [OH⁻]=10⁻ᵖᴼᴴ=10⁻⁹=1.0×10⁻⁹ M.
$[\text{OH}^-]=1.0\times10^{-9},\text{M}$. Apply [OH⁻]=10⁻ᵖᴼᴴ=10⁻⁹=1.0×10⁻⁹ M.
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What is the formula that relates $\text{pH}$ to $[\text{H}^+]$?
What is the formula that relates $\text{pH}$ to $[\text{H}^+]$?
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$\text{pH}=-\log[\text{H}^+]$. Negative log converts hydrogen ion concentration to pH scale.
$\text{pH}=-\log[\text{H}^+]$. Negative log converts hydrogen ion concentration to pH scale.
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What is the relationship between $\text{pH}$ and $\text{pOH}$ at $25^\circ\text{C}$?
What is the relationship between $\text{pH}$ and $\text{pOH}$ at $25^\circ\text{C}$?
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$\text{pH}+\text{pOH}=14.00$. At 25°C, the sum always equals 14 due to water's ion product.
$\text{pH}+\text{pOH}=14.00$. At 25°C, the sum always equals 14 due to water's ion product.
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What is the $\text{pOH}$ when $\text{pH}=3.25$ at $25^\circ\text{C}$?
What is the $\text{pOH}$ when $\text{pH}=3.25$ at $25^\circ\text{C}$?
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$\text{pOH}=10.75$. Use pH+pOH=14.00; pOH=14.00-3.25=10.75.
$\text{pOH}=10.75$. Use pH+pOH=14.00; pOH=14.00-3.25=10.75.
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What is the $\text{pH}$ when $\text{pOH}=1.60$ at $25^\circ\text{C}$?
What is the $\text{pH}$ when $\text{pOH}=1.60$ at $25^\circ\text{C}$?
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$\text{pH}=12.40$. Use pH+pOH=14.00; pH=14.00-1.60=12.40.
$\text{pH}=12.40$. Use pH+pOH=14.00; pH=14.00-1.60=12.40.
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What is the formula that relates $\text{pOH}$ to $[\text{OH}^-]$?
What is the formula that relates $\text{pOH}$ to $[\text{OH}^-]$?
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$\text{pOH}=-\log[\text{OH}^-]$. Negative log converts hydroxide ion concentration to pOH scale.
$\text{pOH}=-\log[\text{OH}^-]$. Negative log converts hydroxide ion concentration to pOH scale.
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What is the pOH of a solution with $[\mathrm{H_3O^+}]=1.0\times10^{-9}\ \mathrm{M}$ at $25^\circ\mathrm{C}$?
What is the pOH of a solution with $[\mathrm{H_3O^+}]=1.0\times10^{-9}\ \mathrm{M}$ at $25^\circ\mathrm{C}$?
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$\mathrm{pOH}=5.00$. $\mathrm{pH}=-\log(1.0\times10^{-9})=9$; $\mathrm{pOH}=14-9=5$.
$\mathrm{pOH}=5.00$. $\mathrm{pH}=-\log(1.0\times10^{-9})=9$; $\mathrm{pOH}=14-9=5$.
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What is $[\mathrm{H_3O^+}]$ for a solution with $\mathrm{pH}=2.50$?
What is $[\mathrm{H_3O^+}]$ for a solution with $\mathrm{pH}=2.50$?
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$[\mathrm{H_3O^+}]=10^{-2.50}=3.2\times10^{-3}\ \mathrm{M}$. Use $[\mathrm{H_3O^+}]=10^{-\mathrm{pH}}=10^{-2.50}$.
$[\mathrm{H_3O^+}]=10^{-2.50}=3.2\times10^{-3}\ \mathrm{M}$. Use $[\mathrm{H_3O^+}]=10^{-\mathrm{pH}}=10^{-2.50}$.
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What is the pH of a $0.015\ \mathrm{M}$ solution of $\mathrm{Ba(OH)_2}$ at $25^\circ\mathrm{C}$?
What is the pH of a $0.015\ \mathrm{M}$ solution of $\mathrm{Ba(OH)_2}$ at $25^\circ\mathrm{C}$?
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$\mathrm{pH}=12.48$. $\mathrm{pH}=14.00-\mathrm{pOH}=14.00-1.52=12.48$.
$\mathrm{pH}=12.48$. $\mathrm{pH}=14.00-\mathrm{pOH}=14.00-1.52=12.48$.
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What is the pOH of a $0.015\ \mathrm{M}$ solution of $\mathrm{Ba(OH)_2}$?
What is the pOH of a $0.015\ \mathrm{M}$ solution of $\mathrm{Ba(OH)_2}$?
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$\mathrm{pOH}=1.52$. $\mathrm{pOH}=-\log(0.030)=-\log(3\times10^{-2})=1.52$.
$\mathrm{pOH}=1.52$. $\mathrm{pOH}=-\log(0.030)=-\log(3\times10^{-2})=1.52$.
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What is the pH of a $2.0\times10^{-5}\ \mathrm{M}$ solution of $\mathrm{HBr}$?
What is the pH of a $2.0\times10^{-5}\ \mathrm{M}$ solution of $\mathrm{HBr}$?
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$\mathrm{pH}=4.70$. $\mathrm{pH}=-\log(2.0\times10^{-5})=5-\log(2)=4.70$.
$\mathrm{pH}=4.70$. $\mathrm{pH}=-\log(2.0\times10^{-5})=5-\log(2)=4.70$.
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What is $[\mathrm{OH^-}]$ for a $0.015\ \mathrm{M}$ solution of $\mathrm{Ba(OH)_2}$?
What is $[\mathrm{OH^-}]$ for a $0.015\ \mathrm{M}$ solution of $\mathrm{Ba(OH)_2}$?
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$[\mathrm{OH^-}]=0.030\ \mathrm{M}$. Ba(OH)₂ releases 2 OH⁻ per unit: $0.015\times^2=0.030\ \mathrm{M}$.
$[\mathrm{OH^-}]=0.030\ \mathrm{M}$. Ba(OH)₂ releases 2 OH⁻ per unit: $0.015\times^2=0.030\ \mathrm{M}$.
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What is the pH of a $4.0\times10^{-2}\ \mathrm{M}$ solution of $\mathrm{NaOH}$ at $25^\circ\mathrm{C}$?
What is the pH of a $4.0\times10^{-2}\ \mathrm{M}$ solution of $\mathrm{NaOH}$ at $25^\circ\mathrm{C}$?
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$\mathrm{pH}=12.60$. $\mathrm{pH}=14.00-\mathrm{pOH}=14.00-1.40=12.60$.
$\mathrm{pH}=12.60$. $\mathrm{pH}=14.00-\mathrm{pOH}=14.00-1.40=12.60$.
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What is the pOH of a $4.0\times10^{-2}\ \mathrm{M}$ solution of $\mathrm{NaOH}$?
What is the pOH of a $4.0\times10^{-2}\ \mathrm{M}$ solution of $\mathrm{NaOH}$?
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$\mathrm{pOH}=1.40$. $\mathrm{pOH}=-\log(4.0\times10^{-2})=2-\log(4)=1.40$.
$\mathrm{pOH}=1.40$. $\mathrm{pOH}=-\log(4.0\times10^{-2})=2-\log(4)=1.40$.
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Identify the pH of pure water at $25^\circ\mathrm{C}$.
Identify the pH of pure water at $25^\circ\mathrm{C}$.
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$\mathrm{pH}=7.00$. Neutral water has $[\mathrm{H_3O^+}]=[\mathrm{OH^-}]=1.0\times10^{-7}\ \mathrm{M}$.
$\mathrm{pH}=7.00$. Neutral water has $[\mathrm{H_3O^+}]=[\mathrm{OH^-}]=1.0\times10^{-7}\ \mathrm{M}$.
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What is the pH of a $3.0\times10^{-4}\ \mathrm{M}$ solution of $\mathrm{Ca(OH)_2}$ at $25^\circ\mathrm{C}$?
What is the pH of a $3.0\times10^{-4}\ \mathrm{M}$ solution of $\mathrm{Ca(OH)_2}$ at $25^\circ\mathrm{C}$?
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$\mathrm{pH}=10.78$. $[\mathrm{OH^-}]=2\times^3.0\times10^{-4}=6.0\times10^{-4}$; $\mathrm{pOH}=3.22$; $\mathrm{pH}=10.78$.
$\mathrm{pH}=10.78$. $[\mathrm{OH^-}]=2\times^3.0\times10^{-4}=6.0\times10^{-4}$; $\mathrm{pOH}=3.22$; $\mathrm{pH}=10.78$.
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State the expression for $K_w$ and its value at $25^\circ\mathrm{C}$.
State the expression for $K_w$ and its value at $25^\circ\mathrm{C}$.
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$K_w=[\mathrm{H_3O^+}][\mathrm{OH^-}]=1.0\times10^{-14}$. Water autoionization constant at $25°\mathrm{C}$.
$K_w=[\mathrm{H_3O^+}][\mathrm{OH^-}]=1.0\times10^{-14}$. Water autoionization constant at $25°\mathrm{C}$.
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Identify the assumption used for strong acids when finding $[\mathrm{H_3O^+}]$.
Identify the assumption used for strong acids when finding $[\mathrm{H_3O^+}]$.
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Strong acids dissociate $\approx100%$: $[\mathrm{H_3O^+}]\approx$ acid molarity. Complete dissociation means initial acid concentration equals $[\mathrm{H_3O^+}]$.
Strong acids dissociate $\approx100%$: $[\mathrm{H_3O^+}]\approx$ acid molarity. Complete dissociation means initial acid concentration equals $[\mathrm{H_3O^+}]$.
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Identify the assumption used for strong bases when finding $[\mathrm{OH^-}]$.
Identify the assumption used for strong bases when finding $[\mathrm{OH^-}]$.
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Strong bases dissociate $\approx100%$: $[\mathrm{OH^-}]\approx$ base molarity $\times$ stoichiometry. Complete dissociation; multiply by OH⁻ per formula unit.
Strong bases dissociate $\approx100%$: $[\mathrm{OH^-}]\approx$ base molarity $\times$ stoichiometry. Complete dissociation; multiply by OH⁻ per formula unit.
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What is $[\mathrm{H_3O^+}]$ for a $0.030\ \mathrm{M}$ solution of $\mathrm{HCl}$?
What is $[\mathrm{H_3O^+}]$ for a $0.030\ \mathrm{M}$ solution of $\mathrm{HCl}$?
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$[\mathrm{H_3O^+}]=0.030\ \mathrm{M}$. HCl is monoprotic strong acid, so $[\mathrm{H_3O^+}]$ equals molarity.
$[\mathrm{H_3O^+}]=0.030\ \mathrm{M}$. HCl is monoprotic strong acid, so $[\mathrm{H_3O^+}]$ equals molarity.
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What is the pH of a $1.0\times10^{-3}\ \mathrm{M}$ solution of $\mathrm{HNO_3}$?
What is the pH of a $1.0\times10^{-3}\ \mathrm{M}$ solution of $\mathrm{HNO_3}$?
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$\mathrm{pH}=3.00$. $\mathrm{pH}=-\log(1.0\times10^{-3})=3.00$ for strong acid HNO₃.
$\mathrm{pH}=3.00$. $\mathrm{pH}=-\log(1.0\times10^{-3})=3.00$ for strong acid HNO₃.
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State the formula for pH in terms of hydronium concentration.
State the formula for pH in terms of hydronium concentration.
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$\mathrm{pH}=-\log[\mathrm{H_3O^+}]$. Negative log converts small $[\mathrm{H_3O^+}]$ values to manageable pH scale.
$\mathrm{pH}=-\log[\mathrm{H_3O^+}]$. Negative log converts small $[\mathrm{H_3O^+}]$ values to manageable pH scale.
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State the formula for pOH in terms of hydroxide concentration.
State the formula for pOH in terms of hydroxide concentration.
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$\mathrm{pOH}=-\log[\mathrm{OH^-}]$. Negative log converts small $[\mathrm{OH^-}]$ values to manageable pOH scale.
$\mathrm{pOH}=-\log[\mathrm{OH^-}]$. Negative log converts small $[\mathrm{OH^-}]$ values to manageable pOH scale.
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What is the relationship between pH and pOH at $25^\circ\mathrm{C}$?
What is the relationship between pH and pOH at $25^\circ\mathrm{C}$?
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$\mathrm{pH}+\mathrm{pOH}=14.00$. At $25°\mathrm{C}$, this sum equals $-\log(K_w)=14.00$.
$\mathrm{pH}+\mathrm{pOH}=14.00$. At $25°\mathrm{C}$, this sum equals $-\log(K_w)=14.00$.
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What is the pH of a solution with $[\mathrm{OH^-}]=1.0\times10^{-12}\ \mathrm{M}$ at $25^\circ\mathrm{C}$?
What is the pH of a solution with $[\mathrm{OH^-}]=1.0\times10^{-12}\ \mathrm{M}$ at $25^\circ\mathrm{C}$?
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$\mathrm{pH}=2.00$. $\mathrm{pOH}=-\log(1.0\times10^{-12})=12$; $\mathrm{pH}=14-12=2$.
$\mathrm{pH}=2.00$. $\mathrm{pOH}=-\log(1.0\times10^{-12})=12$; $\mathrm{pH}=14-12=2$.
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