Question

Calculate the pH of each of the following strong acid solutions (Remember in dilution M1V1 = M2V2 for part (b)). (a)0.0526 M HNO₃

(b)6.00 mL of 1.00 M HCl diluted to 0.650 L

(c) (This is a tricky one that brings the students to my office, Ha Ha)A mixture formed by adding 30.0 mL of 0.018 M HCl to 30.0 mL of 0.015 M HI

Answer 1

(a)0.0526 M HNO₃

HNO₃------->H⁺ + NO₃^-

[HNO₃]=[H⁺]=0.0526

pH=-log[H⁺]=-log(0.0526)=1.279

(b)6.00 mL of 1.00 M HCl diluted to 0.650 L

[HCl]=[H⁺]=9.23*10^-3

pH=-log[H⁺]=-log(9.23*10^-3)=2.03

c)A mixture formed by adding 30.0 mL of 0.018 M HCl to 30.0 mL of 0.015 M HI

net mole of H⁺=mole of HCl+mole of HI=M₁V₁+M₂V₂=(30*10^-3*0.018)+(30*10^-3*0.015)=9.9*10^-4 mole

volume of solution=V₁+V₂=60*10^-3 L

[H⁺]=net mole/net volume=9.9*10^-4/60*10^-3=0.0165 M

pH=-log[H⁺]=-log(0.0165)=1.7825