1. Calculate the pH of a buffer solution made by adding 5.0g of anhydrous sodium

Question

1. Calculate the pH of a buffer solution made by adding 5.0g of anhydrous sodium acetate (NaC2H3O2) to 50mL of 0.100 M acedtic acid. Assume there is no change in volume on adding the salt to the acid. pKa for acetic acid is 4.7.

Buffer capacity refers to the amount of acid or base a buffer can absorb without a significant pH change. A 0.5M buffer can absorb 5x as much acid or base as a 0.1M buffer for a given pH chnage.

You are given 60mL of 0.50M phosphate buffer, pH= 6.83. The starting composition of the buffer, both in terms of concentration and molar quantity of two major phosphate species is:
Concentration of HPO4^2-= 0.304M
Concentration of H2PO4-= 0.196M
Molar quantity of HPO4^2-= 18.2 mmol
Molar quantity of H2PO4-= 11.8 mmol

a) You add 1.7 mL of 1.00 M HCL to the buffer. Calculate the molar quantity of H30+ added as HCL and the final molar quantity of HPO4^2- and H2PO4- at equilibrium?

b) What is the new HPO4^2-/H2PO4- ratio and new pH of solution? The pKa of H2PO4- is 6.64. Use henderson-hasselbach.

C) Now take a fresh 60mL of the 0.50M pH of 6.83 buffer and add 3.7mL of 1.00M NaOH. Caclulate the new pH of the solution.

Explanation / Answer

Part 1

You can calculate the pH of this buffer solution by calculating the concentration of sodium acetate and then applying the Henderson-Hasselbalch equation.

Sodum acetate has a molar mass of 82.03 g/mol. Thus, we can find the moles of sodium acetate added by dividing the mass used by the molar mass:

5.0 g sodium acetate * (1 mol / 82.03 g) = 0.061 mol sodium acetate.

Since the volume doesn't change when the salt is added to the acetic acid, we have a molarity of:

[NaC2H3O2] = 0.061 mol / 0.050 L = 1.22 M in sodium acetate.

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Now we apply Henderson-Hasselbalch:

pH = pKa + log([NaC2H3O2]/[HC2H3O2])

pH = 4.70 + log(1.22 / 0.100 M) = 5.79

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Part 2a

All that was told to you was how many mmol of acid (H2PO4- ) and mmol of conjugate base (HPO42-) you have. When you add HCl to this buffer, the acid will react with the conjugate base and convert it into the weak acid. So, let's calculate how many mmol of HCl we added:

1.7 mL HCl * (1.00 mmol HCl / 1 mL HCl) = 1.7 mmol of HCl added.

Now, let's set up a table to figure out how the number of mmol of each species has changed. Remember, the HCl reacts with HPO42- to make H2PO4-:

So at equilibrium, the new molar quantities are 16.5 mmol of HPO42- annd 13.5 mmol of H2PO4-.

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Part 2b

The ratio of conjugate base to acid is [HPO42-] / [H2PO4-] = 16.5 mmol / 13.5 mmol = 1.22.

Now, we apply Henderson-Hasselbalch, which is:

pH = pKa + log[[HPO42-] / [H2PO4-])

Notice we were given the pKa and we already know that the ratio inside the log is 1.22. So:

pH = 6.64 + log(1.22) = 6.73

This answer makes sense. You added acid to the buffer, and although the buffer does in fact resist the pH change, the pH got lowered by a little bit from what it was.

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Part 2c

The quantities are all the same that they are in part 2b. Now we're simply adding NaOH. Let's calculate the mmol:

3.7 mL NaOH * (1.00 mmol NaOH / 1 mL NaOH) = 3.7 mmol NaOH.

The NaOH reacts with the H2PO4- to make HPO42-

Now we apply Henderson-Hasselbalch again:

pH = 6.64 + log (21.9 / 8.1) = 7.07

HPO42- HCl H2PO4- mmol initial 18.2 mmol 1.7 mmol 11.8 mmol mmol change -1.7 mmol -1.7 mmol +1.7 mmol mmol new 16.5 mmol 0 13.5 mmol

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