Question

Part A) 1.23g of Ba(OH)2 is dissolved in enough pure water at 25 degrees C to make a 50.0mL solution. What is the pH of the solution?
Part B) the 50.0mL solution of Ba(OH)2 in Part A is titrated with a 0.287mol/L solution of hydrochloric acid (HCl). what is the pH after the addition of 45.0mL of HCl?

Answer 1

A) V = 50 mL = 0.05 L

M(Ba(OH)₂) = 171.34 g/mol

m(Ba(OH)₂) = 1.23 g

n = m/M

n(Ba(OH)₂) $= \frac{1.23}{171.34} = 0.007178 \;mol$

[Ba(OH)₂)] $= \frac{0.007178 \;mol}{0.05 \;L} = 0.143 \;M$

Ba(OH)₂ → Ba²⁺ + 2OH^-

Ba(OH)₂ is a strong base

[OH^-] = 2[Ba(OH)₂)]

[OH^-] = $2 \times 0.143 \;M = 0.286 \;M$

[H⁺] $= \frac{1.0 \times 10^{-14}}{[OH-]}$ at 25 degrees C

[H⁺] $= \frac{1.0 \times 10^{-14}}{[0.286]} = 3.4965 \times 10^{-14} \;M$

pH = -log[H⁺]

pH = $-log(3.4965 \times 10^{-14})$

pH = 13.45

B) [HCl] = 0.287 mol/L

V(HCl) = 45 mL = 0.045 L

n(HCl) $= 0.287 \;mol \times 0.045 \;L = 0.012915 \;mol$

[Ba(OH)₂)] = 0.143 M

V(Ba(OH)₂) = 50 mL = 0.05 L

n(Ba(OH)₂) $= 0.143 \; mol/L \times 0.05 \;L = 0.00715 \;mol$

n(OH^-) = 2n(Ba(OH)₂)) $= 2 \times 0.00715 \;mol = 0.0143 \;mol$

n(H⁺) = 0.012915 mol

Remaining n(OH^-) = 0.0143 – 0.012915 = 0.001385 mol

Total V = 0.05 L + 0.45 L = 0.095 L

[OH^-] $= \frac{0.001385 \;mol}{0.095 \;L} = 0.01457 \; M$

[H⁺] $= \frac{1.0 \times 10^{-14}}{[OH-]}$

[H⁺] $= \frac{1.0 \times 10^{-14}}{[0.01457]} = 68.6341 \times 10^{-14} \;M$

pH = -log[H⁺]

pH $= -log(68.6341 \times 10^{-14})$

pH = 12.16