Question

A titration starts off with a 0.1 L aliquot of 0.50 M propionic acid, C3H5O2H. 1. Calculate the pH of this solution. 2. Calculate the pH after adding 10mL of 1.0 M NaOH. -How many moles of base were added to the solution? What effect are they going to have on the acidic solution? By what chemical equation can this effect be explained? -Setup an ice table to calculate the new concentration of hydronium ions after the addition of the base -As an alternative to b now use the Henderson-Hasselbalch equation to calculate the new concentration of hydronium ions after the addition of the base

Show ALL work.

Answer 1

volume of acid = 0.1L = 100mL

[acid] = 0.50M

pKa of ppropionic acid = 4.88

Q1)

Ph of this solution

pH of weak acid = 1/2[pKa -logC]

= 1/2[ 4.88 -log 0.5]

=2.59

Q2) after adding 10mL of 1.0M NaOH

a) moles of NaOH added = V(L) x molarity = 10x10^-3 L x 1 = 1.0x10^-2 moles

b)effect on solution

HA + OH- --------------------------> A- + H2O

100x0.5= 50 0 0 0 initial mmoles

- 10x1.0 = 10 - - change

40 0 10 - equilibrium mmoles

Thus the solution is a buffer and its pH is givenby Hendersen equation as

pH = pKa + log [conjugate bae]/[acid]

= 4.88 + log 10/40

=4.277

Thus the new [H3O+] = antilog -4.277

= 5.284x10^-5 M