Question

Using a 0.25 M phosphate buffer with a pH of 6.2, you add 0.70 mL of 0.50 M HCl to 49 mL of the buffer. What is the new pH of the solution?

Answer 1

moles of Phosphate buffer

pH= PKa+ log [Sat/acid]

6.2 =6.86 + log [conjugate base/acid]

[conjugate base/acid] =0.218776

Conjugate base =0.218776* [acid]

But [Conjugate base ] + acid =0.25

1.218776[acid] =0.25

acid =0.25/1.218776 =0.205 M

conjugate base =0.25-0.205= 0.045 M

Moles of acid = 0.205*49/1000 =0.010045 M

moes of base= 0.045*49/1000 =0.002205 M

Moles of HCl added = 0.5*0.7/1000=0.00035 moles

this neutralizes 0.00035 moles of base, hence moles of base = 0.002205- 0.00035=0.001855

Moles of acid = 0.010045+0.00035=0.010395

Volume of final mixture = 49+0.7 =49.7 ml

Molarity of acid = 0.010395*1000/49.7=0.209155

molarity of base = 0.001855*1000/49.7=0.037324

pH= pKa+log [conjugate base/acid]

pH= =6.86+log [0.037324/0.209155)=6.86-0.74848= 6.11