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Chapter 15: Problem 37

Calculate the mass of sodium acetate that must be added to \(500.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) acetic acid to form a \(\mathrm{pH}=5.00\) buffer solution.

Short Answer

Expert verified
To create a buffer solution with pH=5.00 using 500.0 mL of 0.200M acetic acid, you need to add approximately 14.93 grams of sodium acetate. This is calculated using the Henderson-Hasselbalch equation in conjunction with the pKa of acetic acid and the desired [A-]/[HA] ratio.

Step by step solution

01

Find the pKa of the acetic acid

The pKa is the dissociation constant for the acid, which can be calculated using the formula:pKa = -log(Ka)The Ka value (acid dissociation constant) for acetic acid is \(1.8 × 10^{-5}\). We plug this value into the formula to find the pKa:pKa = -log(\(1.8 × 10^{-5}\)) ≈ 4.74
02

Find the ratio [A-]/[HA] using the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is: pH = pKa + log(\([A^-]/[HA]\)) We know the target pH is 5.00 and we found the pKa to be 4.74. We will now plug these values into the equation and solve for the [A-]/[HA] ratio: 5.00 = 4.74 + log(\([A^-]/[HA]\))
03

Solve for the [A-]/[HA] ratio

We now solve the equation for the ratio, following these steps: 1. Subtract pKa from pH: 5.00 - 4.74 = 0.26 2. Calculate the antilog of 0.26 (also known as the inverse of the logarithm), equal to 10^0.26: \([A^-]/[HA]\) = 10^0.26 ≈ 1.82
04

Calculate the moles of [HA] and [A-] needed

First, we find the initial moles of acetic acid [HA] in the solution: Given that the concentration of acetic acid in the solution is 0.200M and the total volume is 500.0 mL, we convert the volume to liters and multiply by the concentration: moles of [HA] = 0.5 L × 0.200 mol/L = 0.100 mol Now we find the moles of [A-] (sodium acetate) needed for the buffer solution: moles of [A-] = moles of [HA] × \([A^-]/[HA]\) ratio from Step 3 = 0.100 mol × 1.82 ≈ 0.182 mol
05

Calculate the mass of sodium acetate

Now, we can calculate the mass of sodium acetate required for a buffer solution with pH=5.00. The molar mass of sodium acetate (NaC₂H₃O₂) is approximately 82.03 g/mol. We will use the moles of sodium acetate found in step 4: mass of NaC₂H₃O₂ = moles × molar mass = 0.182 mol × 82.03 g/mol ≈ 14.93 g So, approximately 14.93 grams of sodium acetate should be added to the 500.0 mL of 0.200M acetic acid solution to create a buffer solution with pH=5.00.

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Most popular questions from this chapter

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 student titrates an unknown weak acid, HA, to a pale pink phenolphthalein end point with \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\). The student then adds \(13.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\). The \(\mathrm{pH}\) of the resulting solution is \(4.7\). How is the value of \(\mathrm{p} K_{\mathrm{a}}\) for the unknown acid related to \(4.7 ?\)

Consider a buffer solution where [weak acid] \(>\) [conjugate base]. How is the \(\mathrm{pH}\) of the solution related to the \(\mathrm{p} K_{\mathrm{a}}\) value of the weak acid? If [conjugate base] > [weak acid], how is pH related to \(\mathrm{P} K_{\mathrm{a}}\) ?

Consider the titration of \(80.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\), by \(0.400\) \(M\) HCl. Calculate the pH of the resulting solution after the following volumes of \(\mathrm{HCl}\) have been added. a. \(0.0 \mathrm{~mL}\) d. \(40.0 \mathrm{~mL}\) b. \(20.0 \mathrm{~mL}\) e. \(80.0 \mathrm{~mL}\) c. \(30.0 \mathrm{~mL}\)

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