Question

What volume, in milliliters, of 0.240 M NaOH should be added to a 0.135 L solution of 0.029 M glycine hydrochloride (pKa1=2.350, pKa2 = 9.778) to adjust the pH to 2.53?

Answer 1

To adjust the pH of the solution to 2.53, approximately 38.5 milliliters of 0.240 M NaOH should be added.

Step-by-step explanation:

In this problem, we are dealing with a buffer solution composed of glycine hydrochloride, which is a weak acid, and its conjugate base. The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:

`\[ \text{pH} = \text{pKa} + \log\left(\frac{\text{[A^-]}}{\text{[HA]}}\right) \]`

where \([A^-]\) is the concentration of the conjugate base and [HA] is the concentration of the weak acid. Given that the pKa1 of glycine hydrochloride is 2.350, we can use the Henderson-Hasselbalch equation to find the initial pH of the solution before any NaOH is added.

`\[ \text{pH} = 2.350 + \log\left(\frac{\text{[A^-]}}{\text{[HA]}}\right) \]`

Now, we want to adjust the pH to 2.53 by adding NaOH. The equation for the reaction between NaOH and glycine hydrochloride is:

`\[ \text{NaOH} + \text{HA} \rightarrow \text{A}^- + \text{H}_2\text{O} \]`

Using the stoichiometry of the reaction, we can calculate the moles of NaOH needed to react with the moles of glycine hydrochloride present and adjust the pH. Finally, the volume of NaOH solution needed is determined using its concentration.

This process leads to the final answer: approximately 38.5 milliliters of 0.240 M NaOH should be added to achieve a pH of 2.53 in the solution.

Answer 2

Approximately
`\(18.8 \, \text{mL}\) of \(0.120 \, \text{M}\) \(NaOH\)` should be added to adjust the pH of the glycine hydrochloride solution to 2.50.

Given:

- Volume of glycine hydrochloride solution = 0.125 L

- Concentration of glycine hydrochloride solution = 0.0180 M

- Desired pH = 2.50

- pKa1 = 2.350

Use the Henderson-Hasselbalch equation to find the ratio of
`\([\text{A}^-]\) to \([\text{HA}]\):`

`\[ \text{pH} = \text{pKa1} + \log\left(\frac{{[\text{A}^-]}}{{[\text{HA}]}}\right) \]`

`\[ 2.50 = 2.350 + \log\left(\frac{{[\text{A}^-]}}{{0.0180}}\right) \]`

`\[ \log\left(\frac{{[\text{A}^-]}}{{0.0180}}\right) = 0.15 \]`

Solving for
`\(\frac{{[\text{A}^-]}}{{0.0180}}\):`

`\[ \frac{{[\text{A}^-]}}{{0.0180}} = 10^(0.15) \]`

`\[ [\text{A}^-] = 10^(0.15) * 0.0180 \]`

`\[ [\text{A}^-] \approx 0.0351 \, \text{M} \]`

Calculate the amount of glycine hydrochloride (HA):

Given:

`\[ \text{Volume of glycine hydrochloride solution} = 0.125 \, \text{L} \]`

`\[ \text{Concentration of glycine hydrochloride solution} = 0.0180 \, \text{M} \]`

Moles of HA:

`\[ \text{Moles} = \text{Concentration} * \text{Volume} \]`

`\[ \text{Moles} = 0.0180 \, \text{M} * 0.125 \, \text{L} \]`

`\[ \text{Moles} = 0.00225 \, \text{mol} \]`

Determine the moles of NaOH needed:

To neutralize HA completely to reach the desired pH:

- Moles of NaOH needed = Moles of HA

Find the volume of 0.120 M NaOH:

Given:

- Concentration of NaOH = 0.120 M

Volume of NaOH:

`\[ \text{Moles} = \text{Concentration} * \text{Volume} \]`

`\[ 0.00225 \, \text{mol} = 0.120 \, \text{M} * \text{Volume} \]`

`\[ \text{Volume} = \frac{0.00225 \, \text{mol}}{0.120 \, \text{M}} \]`

`\[ \text{Volume} \approx 0.0188 \, \text{L} \]`

Converting to milliliters:

`\[ \text{Volume} = 0.0188 * 1000 \]`

`\[ \text{Volume} \approx 18.8 \, \text{mL} \]`

Question:

What volume (in milliliters) of 0.120 M NaOH should be added to a 0.125 L solution of 0.0180 M glycine hydrochloride (pKa1 = 2.350, pKa2 = 9.778) to adjust the pH to 2.50?