Question

D | Question 6 8 pts Based on the results of your last two problems, determine the pH of 150.0 mL of a 0.50 M CsH5N/ 0.40 MC:HNH' buffer solution after addition of 45.0 miL of 0.75 M HNO3. Ka = 5.9 x 10-6 Hint: Use the Henderson-Hasselbalch equation. IA 1 THAl No new data to save. Last checked at 8:56pm Submit Quiz

Answer 1

Moles = Molarity x Volume (L)

Moles of C5H5N = 0.500 M x (150/1000) L = 0.075 moles
Moles of C5H5NH⁺ = 0.400 M x (150/1000) L = 0.060 moles

Moles of HNO3 = 0.75 M x (45/1000) L = 0.03375 moles

When 45.0 mL of 0.75 M HNO₃ is added;
0.03375 moles of HNO3 will react with 0.03375 moles of C5H5N to produce 0.03375 moles of C5H5NH⁺.

So, moles after addition of HNO3 are
Moles of C5H5N remain = 0.075 mol – 0.03375 mol = 0.04125 mole
Moles of C5H5NH⁺ remain = 0.060 mol + 0.03375 mol = 0.09375 mole

Total volume = 150 mL + 45 mL = 195 mL = 0.195 L

So, concentration terms after addition
[C5H5N] = 0.04125 mol / 0.195 L = 0.212 M
[ C5H5NH⁺] = 0.09375 mol / 0.195 L = 0.481 M

Now, using Henderson-Hesselbalach equation

pH = pKa + log { [salt] \ [acid] }

and

pOH = pKb + log { [salt] \ [bade] }

Kb = 5.9 x 10^-6

pKb = - log Kb = - log (5.9 x 10^-6) = 5.23

Now,

pOH = 5.23+ log (0.481 / 0.212)

pOH = 5.23+ log (0.3558)

pOH = 5.23 - 0.45

pOH = 4.78

pH = 14 – pOH
     = 14 – 4.78
     = 9.22