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Titration and Buffers (5A) Practice Test
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Q1
A 50.0 mL sample of 0.100 M acetic acid ($\mathrm{HA}$) is titrated with 0.100 M NaOH. The reaction is $\mathrm{HA + OH^- \rightarrow A^- + H_2O}$, and $pK_a(\mathrm{HA})=4.76$ (25°C). At the equivalence point, essentially all initial $\mathrm{HA}$ has been converted to $\mathrm{A^-}$, and the pH is determined primarily by base hydrolysis: $\mathrm{A^- + H_2O \rightleftharpoons HA + OH^-}$. For acetate, $K_b=K_w/K_a$ with $K_w=1.0\times10^{-14}$.
Which statement best explains why the pH at the equivalence point is expected to be greater than 7?
Constants provided: $pK_a=4.76$; $K_w=1.0\times10^{-14}$.
A 50.0 mL sample of 0.100 M acetic acid ($\mathrm{HA}$) is titrated with 0.100 M NaOH. The reaction is $\mathrm{HA + OH^- \rightarrow A^- + H_2O}$, and $pK_a(\mathrm{HA})=4.76$ (25°C). At the equivalence point, essentially all initial $\mathrm{HA}$ has been converted to $\mathrm{A^-}$, and the pH is determined primarily by base hydrolysis: $\mathrm{A^- + H_2O \rightleftharpoons HA + OH^-}$. For acetate, $K_b=K_w/K_a$ with $K_w=1.0\times10^{-14}$.
Which statement best explains why the pH at the equivalence point is expected to be greater than 7?
Constants provided: $pK_a=4.76$; $K_w=1.0\times10^{-14}$.