Calculate the pH of each solution. A. 0.18 M CH3NH2
B. 0.18 M CH3NH3Cl
C. a mixture of 0.18 M CH3NH2 and 0.18 M CH3NH3Cl

See Explanations

Explanation:

pH =-log[H₃O⁺] = -log[H⁺]

pOH = -log[OH⁻]

For weak acids [H⁺] = SqrRt(Ka·[Acid])

For weak bases [OH⁻] = SqrRt(Kb·[Base])

pH + pOH = 14

A. Given 0.18M CH₃NH₂; Kb = (4.4 x 10⁻⁴)* => pH = 11.95

CH₃NH₂ + H₂O => CH₃NH₃OH ⇄ CH₃NH₃⁺ + OH⁻;

[OH⁻]  = SqrRt(Kb·[weak base]) = SqrRt(4.4 x 10⁻⁴ x 0.18)M = 8.97 x 10⁻³M

=> pOH = -log[OH⁻] = -log(8.93x10⁻³) = -(-2.05) = 2.05

=> pH = 14 - pOH = 14 - 2.05 = 11.95.

*Kb values for most ammonia derivatives in water can be found online by searching 'Kb-values for weak bases'. Kb-values for methyl amine and methylammonium chloride are both 4.4x10⁻⁴.

B. Given 0.18M CH₃NH₃Cl

In water ... CH₃NH₃Cl => CH₃NH₃⁺ + Cl⁻; Kb(CH₃NH₃Cl) = 4.4 x 10⁻⁴

Cl⁻ + H₂O => No Rxn (i.e.; no hydrolysis occurs) ... Cl⁻ does not react with H₂O.

Hydrolysis Reaction of Methylammonium Ion:

CH₃NH₃⁺ + H₂O => CH₃NH₄OH ⇄ CH₃NH₄⁺ + OH⁻

Ka' x Kb = Kw => Ka' = Kw/Kb = 10⁻¹⁴/4.4 x 10⁻⁴ = 2.27 x 10⁻¹¹                                   Ka' = [CH₃NH₄⁺][OH⁻]/[CH₃NH₄OH] = (x)(x)/(0.18M) = (x²/0.18M) = 2.27 x 10⁻¹¹ => x = [OH⁻] = SqrRt(2.27x10⁻¹¹ x 0.18)M = 2.02 x 10⁻⁶M => pOH = -log(2.02 x 10⁻⁶) = -(-5.69) = 5.69 => pH = 14 - pOH = 14 - 5.69 = 8.31.

*note => the general nature of halide interactions would increase acidity (lower pH) of the halogenated compound.

C. A mixture of 0.18M CH₃NH₂ and 0.18M CH₃NH₃Cl          

Mixture of 0.18M CH₃NH₂ + 0.18M CH₃NH₃Cl

In Water ...

=> 0.18M CH₃NH₃OH + 0.18M CH₃NH₃Cl

=> 0.18M CH₃NH₃⁺ + 0.1M OH⁻ + 0.18M CH₃NH₃⁺ + 0.18M Cl⁻

=> 0.36M CH₃NH₃⁺ + 0.18M OH⁻ + 0.18M Cl⁻

Ka'(CH₃NH₃⁺) x Kb(CH₃NH₂) = Kw => Ka'(CH₃NH₃⁺) = Kw/Kb(CH₃NH₂)

=> Ka'(CH₃NH₃⁺) = (10⁻¹⁴/4.4x10⁻⁴) = 2.27x10⁻¹¹

From the 0.36M CH₃NH₃⁺

=>       CH₃NH₃⁺ + H₂O  ⇄ CH₃NH₄⁺ + OH⁻

C(eq)   0.36M                      x             x     (<= at equilibrium after mixing)

Ka'(CH₃NH₃⁺) = [CH₃NH₄⁺][OH⁻]/[CH₃NH₃⁺] = x²/(0.36M)

=> x = [OH⁻] = SqrRt(Ka'(CH₃NH₃⁺)·0.36M) = SqrRt(2.27x10⁻¹¹/0.36) = 0.0126M

=> Total [OH⁻] = 0.0126M + 0.18M = 0.1926M from hydrolysis process

=> final solution mix is therefore, 0.1926M in OH⁻ + 0.18M in Cl⁻

∴pH = -log[H⁺] = -log(5.192x10⁻¹⁴) = -(-13.29) = 13.29 for solution mix

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the political sponsor of weights and measures reform in the french revolutionary national assembly was the bishop of autun, better known as talleyrand. under his auspices, the french academy appointed several committees to carry out the work of developing a usable system of weights and measures for france. one of the committees recommended a decimalized measurement system based upon a length equal to one ten-millionth of the length of a quadrant of the earth’s meridian (i.e., one ten-millionth of the distance between the equator and the north pole).