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 12.081.92
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 2.46Explanation / 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
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