chemistry homework questions

Question

You mix 200 mL of 0.3 M NH₃ (pKb = 4.75) with 300 mL of 0.3 M NH₄Cl to make 500 mL of buffer solution. If you then wanted to adjust the pH to be 8.7, how many moles of solid HBr should you add? Hint 1: you can convince yourself that you need to add some HBr by first calculating the initial buffer's pH, which you'll find is 9.07. Hint 2: If you know the pH you want, you know what you want the equilibrium [OH⁻] to be. Hint 3: This is a situation where it may be helpful to define another variable to help you write down the "initial" row in an ICE table.

Answer 1

First, find the initial pH:

Moles NH₃ = 200 ml x 1 L / 1000 ml x 0.3 mol / L = 0.06 moles

Moles NH₄⁺ = 300 ml x 1 L / 1000 ml x 0.3 mol / L = 0.09 moles

Final [NH₃] = 0.06 mol / 0.5 L = 0.12 M

Final [NH₄⁺] = 0.09 mol / 0.5 L = 0.18 M

Henderson Hasselbalch equation for a basic buffer:

pOH = pKb + log [conj.acid] / [base]

pOH = 4.75 + log [0.18/0.12]

pOH = 4.75 + 0.18

pOH = 4.93

pH = 14 - pOH

pH = 9.07 = Initial pH

So, to get the pH to be 8.7, we will have to add acid (H⁺).

We can use the HH equation again, using the desired pOH and solving for the ratio of [NH₄⁺] / [NH₃].

The desired pOH will be 14 - desired pH = 14 - 8.7

Desired pOH = 5.3

pOH = pKb + log [NH₄⁺] / [NH₃]

5.3 = 4.75 + log [NH₄⁺] / [NH₃]

log [NH₄⁺] / [NH₃] = 0.55

[NH₄⁺] / [NH₃] = 3.55

NH₃ + H⁺ ====> NH₄⁺

0.06.......x.................0.09........Initial

-x..........-x...............+x............Change

0.06-x....0............0.09+x........Equilibrium

0.09+x / 0.06-x = 3.55

0.213 - 3.55x = 0.09 + x

x = 0.027 = moles HBr needed

(be sure to check all of the math)