Question

A. What is the pH of the following solutions?

i. [H₃O⁺] = 1.0 X 10^-6

ii. [H₃O⁺] = 4.4 X 10^-2

B. What volume in mL of 0.115 M sodium hydroxide would neutralize 25.0 mL of a 0.106 M sulfuric acid solution during a titration?

Answer 1

A)
i)
Given:
[H3O+] = 1*10^-6 M

use:
pH = -log [H3O+]
= -log (1*10^-6)
= 6.00
Answer: 6.00

B)
Given:
[H3O+] = 4.4*10^-2 M

use:
pH = -log [H3O+]
= -log (4.4*10^-2)
= 1.3565
Answer: 1.36

B)
Balanced chemical equation is:
H2SO4 + 2 NaOH ---> Na2SO4 + 2 H2O


Here:
M(H2SO4)=0.106 M
M(NaOH)=0.115 M
V(H2SO4)=25.0 mL

According to balanced reaction:
2*number of mol of H2SO4 =1*number of mol of NaOH
2*M(H2SO4)*V(H2SO4) =1*M(NaOH)*V(NaOH)
2*0.106 M *25.0 mL = 1*0.115M *V(NaOH)
V(NaOH) = 46.1 mL

Answer: 46.1 mL