Calculate the pH of a 0.034 M solution of RbOH. Calculate the pH of a 0.034 M so

Question

Calculate the pH of a 0.034 M solution of RbOH. Calculate the pH of a 0.034 M solution of Sr(OH)_2. Compare your answers for questions 4 and 5. Explain why they differ even though both compounds are hydroxide salts. Calculate the pH of a 0.28 M solution of HNO_2. Calculate the pH of a 0.60 M solution of C_6H_5NH. Calculate the pH of a 0.60 M solution of (CH_3)_2 NH. Compare your answers for questions 8 and 9. Explain why they differ even though both compounds are present at the same concentration.

Explanation / Answer

4.

pH= 14 -log[OH]

[OH] = 0.034

-log[0.034]= 1.47

pH = 14 - 1.47 = 12.53

5.

pH of Sr(OH)2

[OH] = 0.034

pH = 14 - log[OH]

since 2 OH is for Sr(OH)2

[OH] = 2 x 0.034 = 0.068

-log[OH] = -log[0.068] = 1.17

pH = 14 – 1.17 = 12.83

6)

The pH value is greater for Sr(OH)2 compared to Rb(OH) because in two OH groups are present in Sr(OH)2

7.

pH= -log[H+]

[H+] = 0.28 M

pH = -log[ 0.28 ] = 0.552

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