50 ml of H2O is added to 50 ml of 1×10-3Mbarium hydroxide solution. What is the pH of the resulting solution?

V1+V2=100mL,N2=2;N1+M1×n=1×10-3×2=2×10-3N1V1=N2V22×10-3×50=N2×100N2=10-3pOH = 3 pH = 11

50 ml of H2O is added to 50 ml of 1×10-3Mbarium hydroxide solution. What is the pH of the resulting solution?

V1+V2=100mL,N2=2;N1+M1×n=1×10-3×2=2×10-3N1V1=N2V22×10-3×50=N2×100N2=10-3pOH = 3 pH = 11

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50 ml of $H_{2} O$ is added to 50 ml of $1 \times 10^{- 3} M$ barium hydroxide solution. What is the pH of the resulting solution?

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3.0

11.0

3.3

11.7

answer is D.

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Detailed Solution

$V_{1} + V_{2} = 100 mL, N_{2} = 2; N_{1} + M_{1} \times n = 1 \times 10^{- 3} \times 2 = 2 \times 10^{- 3}$

$N_{1} V_{1} = N_{2} V_{2}$

$2 \times 10^{- 3} \times 50 = N_{2} \times 100$

$N_{2} = 10^{- 3}$

pOH = 3

pH = 11

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