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Chapter 15: Problem 36

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

Short Answer

Expert verified
The pH of the solution containing 0.60 M HF and 1.00 M KF is approximately 3.71, using the pKa of HF (\(\approx 3.46\)) and the Henderson-Hasselbalch equation.

Step by step solution

01

Write down the given information

We have a solution containing 0.60 M HF (weak acid) and 1.00 M KF (the source of F-, a conjugate base). Our goal is to calculate the pH of this solution.
02

Find the pKa of HF

The dissociation constant, Ka, of HF is 3.5 × 10^{-4}. To find the pKa, we use the formula: \(pKa = -\log{Ka}\). So, the pKa of HF is given by: \[pKa = -\log{(3.5 × 10^{-4})} \approx 3.46\]
03

Use Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given by: \[pH = pKa + \log{\frac{[A^-]}{[HA]}}\] Where [A-] is the concentration of the conjugate base (F-), and [HA] is the concentration of the weak acid (HF). In our case: \([A^-] = 1.00\,{M}\) (from KF) and \([HA] = 0.60\,{M}\) (HF)
04

Calculate the pH of the solution

Substitute the values of pKa, [A-], and [HA] from Steps 2 and 3 into the Henderson-Hasselbalch equation to find the pH. \[pH = 3.46 + \log{\frac{1.00}{0.60}} \approx 3.46 + 0.25 = 3.71\] So, the pH of the solution containing 0.60 M HF and 1.00 M KF is approximately 3.71.

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Most popular questions from this chapter

no question at book

Phosphate buffers are important in regulating the \(\mathrm{pH}\) of intra- cellular fluids at pH values generally between 7.1 and \(7.2 .\) a. What is the concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}^{2-}\) inintracellular fluid at \(\mathrm{pH}=7.15 ?\) $$ \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-8} $$ b. Why is a buffer composed of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ineffective in buffering the pH of intracellular fluid? $$ \mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{2}=7.5 \times 10^{-3} $$

You make 1.00 \(\mathrm{L}\) of a buffered solution \((\mathrm{pH}=4.00)\) by mixing acetic acid and sodium acetate. You have 1.00\(M\) solutions of each component of the buffered solution. What volume of each solution do you mix to make such a buffered solution?

Derive an equation analogous to the Henderson-Hasselbalch equation but relating pOH and \(\mathrm{p} K_{\mathrm{b}}\) of a buffered solution composed of a weak base and its conjugate acid, such as \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4}^{+}\)

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 \(\mathrm{OH}^{-}\) .
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