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Chapter 15: Problem 28

Calculate the \(\mathrm{pH}\) of a solution that is \(0.60 \mathrm{M} \mathrm{HF}\) and \(1.00 \mathrm{M} \mathrm{KF}\).

Short Answer

Expert verified
The pH of the solution containing 0.60 M HF and 1.00 M KF is approximately 3.39.

Step by step solution

01

Write the chemical equation for the acidic dissociation of HF

First, we need to write the chemical equation for the acidic dissociation of HF in aqueous solution: \[HF \rightleftharpoons H^+ + F^-\]
02

Find the \(K_a\) value for HF

Next, we need to find the acid dissociation constant (\(K_a\)) for HF. We can obtain it from a table of common \(K_a\) values in any chemistry textbook or online source. For HF, the \(K_a\) value is \(6.8\times10^{-4}\).
03

Use the Henderson-Hasselbalch equation

To find the \(\mathrm{pH}\) of the buffer solution, we will use the Henderson-Hasselbalch equation: \[\mathrm{pH} =\mathrm{p}K_a + \log \frac{[\mathrm{F^-}]}{[\mathrm{HF}]}\] where \([\mathrm{F^-}]\) and \([\mathrm{HF}]\) are the molar concentrations of F⁻ and HF, respectively.
04

Plug in the given values and solve for pH

Now plug in the given values of \([\mathrm{F^-}]\), \([\mathrm{HF}]\) and \(K_a\), and calculate the \(\mathrm{pH}\) of the buffer solution: \[ \mathrm{pH} = -\log (6.8\times10^{-4}) + \log \frac{1.0\,\mathrm{M}}{0.60\,\mathrm{M}}\] \[ \mathrm{pH} = 3.17 + \log 1.67\] \[ \mathrm{pH} \approx 3.17 + 0.22\] \[ \mathrm{pH} \approx 3.39 \] So, the \(\mathrm{pH}\) of the solution containing 0.60 M HF and 1.00 M KF is approximately 3.39.

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Most popular questions from this chapter

Consider the following two acids: O=C(O)c1ccccc1O Salicylic acid $\mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}$ Adipic acid $\mathrm{p} K_{\mathrm{a}_{\mathrm{l}}}=4.4 \mathrm{l} ; \mathrm{p} K_{\mathrm{L}_{3}}=5.28$ In two separate experiments the \(\mathrm{pH}\) was measured during the titration of \(5.00 \mathrm{mmol}\) of each acid with $0.200 \mathrm{M} \mathrm{NaOH}$. Each experiment showed only one stoichiometric point when the data were plotted. In one experiment the stoichiometric point was at $25.00 \mathrm{~mL}\( added \)\mathrm{NaOH}$, and in the other experiment the stoichiometric point was at \(50.00 \mathrm{~mL} \mathrm{NaOH}\). Explain these results. (See Exercise 103.)

Sketch two pH curves, one for the titration of a weak acid with a strong base and one for a strong acid with a strong base. How are they similar? How are they different? Account for the similarities and the differences.

The common ion effect for weak acids is to significantly decrease the dissociation of the acid in water. Explain the common ion effect.

Figure \(15.4\) shows the pH curves for the titrations of six different acids by \(\mathrm{NaOH}\). Make a similar plot for the titration of three different bases by \(0.10 M\) HCl. Assume \(50.0 \mathrm{~mL}\) of \(0.20 \mathrm{M}\) of the bases and assume the three bases are a strong base \((\mathrm{KOH})\), a weak base with \(K_{\mathrm{b}}=1 \times 10^{-5}\), and another weak base with \(K_{\mathrm{b}}=1 \times 10^{-10}\)

A certain buffer is made by dissolving \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in some water. Write equations to show how this buffer neutralizes added \(\mathrm{H}^{+}\) and \(\overline{\mathrm{OH}}^{-}\).
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