Chemistry buffer question!

Question

What is the pH of the buffer that results when 12 g sodium formate (NaHCO2) is mixed with 250 mL of 0.50 M formic acid (HCO2H) and diluted with water to 1.0 L? (Kaof HCO2H = 1.8e10-4)Molar mass NaHCO2= 68.01 g/mol

Answer is:3.89

Don't understand how to solve

Answer 1

I must say that the answer provided by @Philip S. is excellent, but I wanted to offer another approach that might help you if you need to find the answer quicker, as on an exam, for example. The previous answer is really good in that it explains exactly what is going on, and you should understand that. But here is an alternate approach...

Become familiar with the Henderson Hasselbalch equation:

pH = pKa + log [salt]/[acid]

pH = ?

pKa = -log Ka = -log 1.8x10^-4 = 3.74

[salt] = [HCO2^-] 12 g x 1 mol/68.01 g = 0.176 moles and since this is in 1 L [salt] = 0.176 M

[acid] = [HCO2H]: (250 ml)(0.5 M) = (1000 ml )(x M) and x = 0.125 M (this is simply M1V1 = M2V2 dilution)

Going back to HH we have...

pH = 3.74 + log (0.176/0.125) = 3.74 + log 1.41

pH = 3.74 + 0.148

pH = 3.89

Answer 2

It's always good to understand what is going on first so that your answer is correct. So let's start with the initial solution (HCO2H without NaHCO2).

Formic Acid (HCO2H) is a weak acid, so it dissolves in water partially.

HCO2H ---> CO2H- + H+

Using our equilibrium constant equation we get:

Ka = [CO2H-][H+] / [HCO2H]

Now, let's do an ICE Chart:

[HCO2H] [CO2H-] [H+]

Initial .5 0 0

Change -x x x

Equilibrium .5-x x x

so:

1.8 x 10^-4 = x² / (.5 - x)

Solving for x using the quadratic equation we get x = .009397

This means our [H+] = .009397, using our pH equation:

pH = -log[H+]

pH = -log (.009397)

pH = 2.03 (This is without adding NaHCO2; I only show this to show the pH is less than your answer and the reason for this is below)

Going back to our chemical equation:

HCO2H ---> CO2H- + H+

If we add NaHCO2, we are essentially are adding CO2H- anions since the salt will dissolve (there is a sodium cation in the compound). Applying Le Chatlier's principle, if we add more of one component to the right, we willbe driving the reaction to make more reactant. This means, if we add NaHCO2 to the HCO2H solution, the pH should go up.

Now recalculate molarities of solution after diluting:

M_HCO2H = .5 M x .25L / 1 L = .125 M_HCO2H

M_NaHCO2 = 12 g NaHCO2 x 1 mol NaHCO2 / 68.01 g NaHCO2 / 1 L = 0.176 M_NaHCO2

Run an ICE Chart again:

[HCO2H] [CO2H-] [H+]

Initial .125 .176 0

Change -x x x

Equilibrium .125-x .176 + x x

Use the Equilibrium constant equation:

Ka = [CO2H-][H+] / [HCO2H]

1.8 x 10^-4 = (.176 + x) * x / (.125 - x)

Solving for x using the quadratic equation we get x = .000127618

Since x represents [H+], [H+] = .000127618

Using our pH equation:

pH = -log[H+]

pH = -log(.000127618)

pH = 3.894