Buffer question

[yui_96]

Can someone help me work out this question? Thank you!

When 2mL of 0.1 M NaOH (aq) is added to 100mL of a solution containing 0.1M HCIO (aq) and 0.1M NaCIO (aq), what type of change in the pH of the solution takes place? (Ka for HCIO is 3.2*10^-8)
A. A slight (<0.1 pH unit) increase
B. A slight (<0.1 pH unit) decrease
C. A significant (>0.1 pH unit) increase
D. A significant (>0.1 pH unit) decrease

The answer given is A and it only says because a strong base is added to a buffer solution the pH change is slight. I'm trying to figure out whether the exact number is less or more than 0.1pH unit but I'm not sure how to approach this.

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[NextStepTutor_1]

Hi @yui_96 -

Using the Henderson-Hasselbalch equation (pH = pKa + log[A-]/[HA]), we can predict that the pH will change proportionally to the change in log[ClO-]/[HClO].

The first step here is to figure out how many moles of NaOH we're dealing with. 2 mL * 0.1 M = 0.2 mmol of OH-. That will deprotonate 0.2 mmol of HClO. Our next task is to figure out what that will do to the ratio of [ClO-]/[HClO]. The solution contains 100 mL * 0.1M = 10 mmol before the OH- is added. Deprotonating 0.2 mmol of that means that we will have 9.8 mmol of HClO and 10.2 mmol of ClO- in 102 mL of solution (100 mL original + 2 mL of the NaOH solution). The concentration of ClO- will now be 0.1 M (0.0102 mol/0.102 L), and the concentration of HClO will be 0.096 M. The ratio of [ClO-]/[HClO] is 0.1/0.096 = 1.04. Log(1.04) = 0.017. Per the Henderson-Hasselbalch equation, this will be our change in pH (b/c pKa is a constant), so the Q is right that the increase will be <0.1 pH unit.

I tried to be quite careful with the math here, and used a calculator -- but you don't need that level of the precision for the MCAT. If you ever had to deal w/ a quantitative question like this, you could take shortcuts by, for instance, neglecting the extra 2 mL of volume in the final calculation concentrations.

Hope this helps & best of luck!

 

[NextStepTutor_1]

P.S., an extra step that I didn't include is that the original log[A-]/[HA] term is 0, because the concentration of ClO- and HClO is equal in the initial setup (their ratio is 1, and log(1) = 0), so the modified log[A-]/[HA] term we calculate after adding the NaOH solution itself is the change in the log[A-]/[HA] term, mathematically speaking.