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1) What is the pH of a solution prepared by diluting 75.00 mL of 0.020 M Ba(OH)

ID: 879429 • Letter: 1

Question

1) What is the pH of a solution prepared by diluting 75.00 mL of 0.020 M Ba(OH)2 with enough water to produce a total volume of 250.00 mL?

1.92

2) Calculate the pH of the following solutions:

0.030 M HClO4

Express your answer using two decimal places.

1.2 M KOH

Express your answer using two decimal places.

0.065 M NaOH

Express your answer using two decimal places.

3) What is the pH of a 0.10 M NaF solution? (Ka of HF = 3.5 × 104)

11.78 2.22 12.08

1.92

2) Calculate the pH of the following solutions:

0.030 M HClO4

Express your answer using two decimal places.

1.2 M KOH

Express your answer using two decimal places.

0.065 M NaOH

Express your answer using two decimal places.

3) What is the pH of a 0.10 M NaF solution? (Ka of HF = 3.5 × 104)

8.23 5.77 2.23 2.46

Explanation / Answer

1. Ba(OH)2 is a strong base and dissociates completely.

Ba(OH)2(aq) -----> Ba^2+(aq) + 2OH^-(aq)
0.020 M ....................................... 2 x 0.020 = 0.040 M

After dilution;
M1V1 = M2V2
(0.040 M)(75.00 mL) = M2(250.00 mL)
M2 = 0.012 mol/L = [OH-]

pOH = -log[OH-] = -log(0.012) = 1.92

pH = 14.00 - 1.92= 12.07