Question

In: Chemistry

1. The pH of a 0.42M dimethylamine solution is: A. 2.5 B.1.80 C.1.7 D.12.2 2.The pH...

1. The pH of a 0.42M dimethylamine solution is:

A. 2.5 B.1.80 C.1.7 D.12.2

2.The pH of a buffer solution composed of 0.020M HClO and 0.014M NaClO is:

A.7.37 B. 7.52 C. 7.68

3.When a base is added to a buffer composed of HF and NaF , the OH- is neutralized by:

A. F- B. HF C. Water

4. calculate the solubility of iron(II) hydroxide at 25oC

A. 5.8 x 10-6
B. 2.8 x 10-8
C. 9.2 x 10-6

5.. The pH of a 0.0145M perchloric acid solution is:

A. 1.84 B. 7.0 C.12.2

6.When 400ml of a 0.01M CaCl solution is mixed with a 300ml if a NaOH solution at pH 9.0, a precipitate will form.

True Or False

Solutions

Expert Solution

1. D.12.2
dimethylamine is base, so its pH will be greater than 7.

2.Choice C. 7.68
The pH of a buffer solution composed of 0.020M HClO and 0.014M NaClO is:
Hypochlorous acid HOCl, Ka = 3E-8
Henderson-Hasselbach equation, pH = pKa + Log([ClO-]/[HClO])= -Log(3E-8) + Log(0.014/0.02)
pH = 8 - 0.15 = 7.85
Choice C. 7.68 is closer.

3. B.
HF is the acid whic can neutralize base OH-


4. C 9.2 x 10-6
calculate the solubility of iron(II) hydroxide at 25oC
Fe(OH)2(s) = Fe(2+) + 2OH(-)
Ksp(Fe(OH)2) = 4.1E-15 = S*(2*S)^2 = 4*S^3
solubility, S = 1.01E-5
Choice C 9.2 x 10-6 is closer

5. A. 1.84
The pH of a 0.0145M perchloric acid solution is:
perchloric acid is strong acid and [H+] = 0.0145
so pH = -Log(H+) = 1.84
          


6.False
400ml of a 0.01M CaCl solution is mixed with a 300ml
of a NaOH solution at pH 9.0,

CaCl2 + NaOH = Ca(OH)2(s) + 2NaCl(aq)
Ca(OH)2(s) = Ca(2+) + 2OH-
Ksp(Ca(OH)2) = 6.5E-6
Qsp(Ca(OH)2) = [Ca(2+)]*[OH-]^2
final [Ca(2+)] = 0.01*400/(400+300) = 0.0057 M
pH=9 of [OH-] = 10^(-14+pH) = 1.0E-5
final [OH-] = 1E-5*300/(400+300) = 0.43E-5 M
Qsp(Ca(OH)2) = [Ca(2+)]*[OH-]^2 =   [0.0057]*[0.43E-5]^2 = 1.04E-13 < Ksp

No ppt as Qsp < Ksp


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