Answer:
12.15, 2.23
Explanation:
Using the dilution formula
C₁V₁ = C₂ V₂
where C₁ = 1.46 M
V₁ = 1.48 ml = 0.00148 L
C₂ = unknown
V₂ = 150 ml + 1.48ml = 151.48 ml = 0.15148L
1.46 M × 0.00148 L = C₂×0.15148L
C₂ = 0.0143 M
pOH = - log ( OH⁻) = - log ( 0.0143) = 1.85
pH + pOH = 14
pH = 14 - 1.85 = 12.15
b) using the same formula
C₁V₁ = C₂ V₂
C₁ = 0.99 M
V₁ = 1.49 ml = 0.00149 L
C₂ = unknown
V₂ = 250 ml + 1.49 ml = 251.49 ml = 0.25149 L
0.99 M × 0.00149 L = C₂ × 0.25149 L
C₂ = 0.00587
pH = - log ( H⁺) = - ( -2.23) = 2.23