A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M NaOH.Determine the pH of the solution after the addition of 200.0 mL of NaOH.The K_a of HF is 3.5 × 10^-4.

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The reaction between HF and NaOH is a neutralization reaction where HF (a weak acid) reacts with NaOH (a strong base) to form water and the fluoride ion (F-). The amount of HF initially present is $0.100\,L \times 0.20\,M = 0.020\,mol$. The amount of NaOH added is $0.200\,L \times 0.10\,M = 0.020\,mol$. Since the moles of HF and NaOH are equal, all the HF will react with NaOH to form F- and water, leaving no HF in the solution. The solution will contain only F- ions and Na+ ions from the NaOH, making it a solution of sodium fluoride (NaF), which is a salt of a weak acid (HF) and a strong base (NaOH).The pH of the solution will be determined by the hydrolysis of the fluoride ion (F-), which is the conjugate base of HF. The hydrolysis equation is:$$ F^- + H_2O \rightleftharpoons HF + OH^- $$The equilibrium constant for this reaction, $K_b$, can be found using the relation $K_w = K_a \times K_b$, where $K_w$ is the ion-product constant for water ($1.0 \times 10^{-14}$) and $K_a$ is the acid dissociation constant for HF ($3.5 \times 10^{-4}$).$$ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{3.5 \times 10^{-4}} = 2.857 \times 10^{-11} $$Given that the concentration of F- is $0.020\,mol$ in a total volume of $0.100\,L + 0.200\,L = 0.300\,L$, the concentration of F- is $\frac{0.020\,mol}{0.300\,L} = 0.0667\,M$.Using the $K_b$ and the concentration of F-, we can find the concentration of OH- produced by the hydrolysis and then calculate the pOH and subsequently the pH of the solution. However, the calculation of pH directly from the given information and the hydrolysis of F- leads to a solution that is basic due to the production of OH- ions. The pH value that corresponds to a basic solution from the given options is 8.14, which is consistent with the expected outcome of a solution containing the conjugate base of a weak acid after neutralization.

A 1000 ML Sample of 0