Question

# 500.0 mL of 0.160 M NaOH is added to 575 mL of 0.250 M weak acid...

500.0 mL of 0.160 M NaOH is added to 575 mL of 0.250 M weak acid (Ka = 2.43 × 10-5). What is the pH of the resulting buffer?

To calculate the pH of the resulting buffer use Henderson - Hasselbalch equation:

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

pKa = - log Ka= - log 2.43 × 10-5= 4.61

Now calculate the concentration of salt and acid as follows:

Mol NaOH in 500mL of 0.160M solution

= 500/1000*0.110

= 0.08 mol NaOH
Mol Weak acid in 575mL of 0.250M solution

= 575/1000*0.250

= 0.144mol

NaOH + HA = NaA+ H2O
0.08 mol          0.08 mol

Here 0.08 mol of the Na salt and 0.

144 - 0.08 = 0.064 mol of unreacted acid remain .

The total volume = 500+575 mL = 1075 ml

= 1.075L
Molarity of salt = 0.08/1.075 = 0.074M
Molarity of acid = 0.064/1.075 = 0.060 M

To calculate the pH of the resulting buffer use Henderson - Hasselbalch equation:

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

pH = 4.61+ log 0.074/0.060

pH = 4.61+0.0910

pH = 4.701

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