Question

Part A) Calculate [OH−] for strong base solution formed by mixing 15.0 mL of 1.00×10−2 M Ba(OH)2 with 34.0 mL of 6.2×10−3 M NaOH.

Part B) Calculate pH for strong base solution formed by mixing 15.0 mL of 1.00×10−2 M Ba(OH)2 with 34.0 mL of 6.2×10−3 M NaOH.

Answer 1

Both A and B are same

A)
Concentration of OH- in mixture = (n1*C1*V1+ n2*C2*V2) / (V1+V2)
where C1 --> Concentration of 1 component
V1-->volume of 1 component
C2 --> Concentration of other component
V2-->volume of other component
n1 --> number of particle from 1 molecule of 1st component
n2 --> number of particle from 1 molecule of 2nd component

Here n1 = 2 as Ba(OH)2 as 2 OH-
Here n2 = 1 as NaOH as 1 OH-
use:
C = (n1*C1*V1+ n2*C2*V2) / (V1+V2)
C = (2*0.01*15+1*0.0062*34)/(15+34)
C = 0.0104 M
Answer: 1.04*10^-2 M