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Chapter 14: Problem 70

Consider the titration of \(\mathrm{HF}\left(K_{\mathrm{a}}=6.7 \times 10^{-4}\right)\) with \(\mathrm{NaOH}\). What is the \(\mathrm{pH}\) when a third of the acid has been neutralized?

Short Answer

Expert verified
Answer: The pH of the solution is approximately 3.05 when one-third of the acid has been neutralized.

Step by step solution

01

Determine the Moles of HF and F- after Neutralization

Let's assume that the initial number of moles of HF is x. After one-third of the moles of HF have been neutralized by NaOH, we will have 2x/3 moles of HF and x/3 moles of F-
02

Setting up the Equation for H+ Concentration

Now, we can set up the equation for the H+ concentration. The acid dissociation constant, Ka, is given by the following expression: $$ K_a = \frac{[\mathrm{H}^{+}][\mathrm{F}^{-}]}{[\mathrm{HF}]} $$ We are given the value of Ka for HF, which is \(6.7 \times 10^{-4}\). From step 1, we know the concentrations of HF and F-. Let's substitute these values into the Ka expression to find the concentration of H+: $$ 6.7 \times 10^{-4} = \frac{[\mathrm{H}^{+}]\frac{x}{3}}{\frac{2x}{3}} $$
03

Solve for H+ Concentration

Now, we can solve the equation for the H+ concentration. $$ [\mathrm{H}^{+}] = 6.7 \times 10^{-4} \times \frac{\frac{2x}{3}}{\frac{x}{3}} = \frac{4}{3}(6.7 \times 10^{-4}) $$ After calculating the value, we get: $$ [\mathrm{H}^{+}] = 8.93 \times 10^{-4} \thinspace \mathrm{M} $$
04

Calculate the pH of the Solution

Finally, we can use the H+ concentration to calculate the pH of the solution using the following formula: $$ \mathrm{pH} = -\log[\mathrm{H}^{+}] $$ Substituting the H+ concentration, we get: $$ \mathrm{pH} = -\log(8.93 \times 10^{-4}) $$ After calculating, the pH of the solution is about 3.05 when one-third of the acid has been neutralized.

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Most popular questions from this chapter

Write a balanced net ionic equation for the reaction of each of the following aqueous solutions with \(\mathrm{H}^{+}\) ions. (a) sodium fluoride (b) barium hydroxide (c) potassium dihydrogen phosphate \(\left(\mathrm{KH}_{2} \mathrm{PO}_{4}\right)\)

A buffer is prepared by dissolving \(0.0250 \mathrm{~mol}\) of sodium nitrite, \(\mathrm{NaNO}_{2}\), in \(250.0 \mathrm{~mL}\) of \(0.0410 \mathrm{M}\) nitrous acid, \(\mathrm{HNO}_{2}\). Assume no volume change after \(\mathrm{HNO}_{2}\) is dissolved. Calculate the \(\mathrm{pH}\) of this buffer.

There is a buffer system \(\left(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}-\mathrm{HPO}_{4}{ }^{2-}\right)\) in blood that helps keep the blood \(\mathrm{pH}\) at about \(7.40 .\left(\mathrm{K}_{\mathrm{a}} \mathrm{H}_{2} \mathrm{PO}_{4}^{-}=6.2 \times 10^{-8}\right)\). (a) Calculate the \(\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right] /\left[\mathrm{HPO}_{4}^{2-}\right]\) ratio at the normal \(\mathrm{pH}\) of blood. (b) What percentage of the \(\mathrm{HPO}_{4}{ }^{2-}\) ions are converted to \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) when the \(\mathrm{pH}\) goes down to \(6.80\) ? (c) What percentage of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ions are converted to \(\mathrm{HPO}_{4}{ }^{2-}\) when the \(\mathrm{pH}\) goes up to \(7.80\) ?

WEB To make a buffer with \(\mathrm{pH}=3.0\) from \(\mathrm{HCHO}_{2}\) and \(\mathrm{CHO}_{2}^{-}\), (a) what must the [HCHO \(\left._{2}\right] /\left[\mathrm{CHO}_{2}^{-}\right]\) ratio be? (b) how many moles of \(\mathrm{HCHO}_{2}\) must be added to a liter of \(0.139 \mathrm{M}\) \(\mathrm{NaCHO}_{2}\) to give this \(\mathrm{pH}\) ? (c) how many grams of \(\mathrm{NaCHO}_{2}\) must be added to \(350.0 \mathrm{~mL}\) of \(0.159 \mathrm{MHCHO}_{2}\) to give this \(\mathrm{pH}\) ? (d) What volume of \(0.236 \mathrm{M} \mathrm{HCHO}_{2}\) must be added to \(1.00 \mathrm{~L}\) of a \(0.500 \mathrm{M}\) solution of \(\mathrm{NaCHO}_{2}\) to give this pH? (Assume that volumes are additive.)

What is the \(\mathrm{pH}\) of a \(0.1500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution if (a) the ionization of \(\mathrm{HSO}_{4}^{-}\) is ignored? (b) the ionization of \(\mathrm{HSO}_{4}^{-}\) is taken into account? \(\left(K_{\mathrm{a}}\right.\) for \(\mathrm{HSO}_{4}^{-}\) is \(\left.1.1 \times 10^{-2} .\right)\)
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