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Andrei [34K]
3 years ago
6

A solution is prepared at 25°C that is initially 0.18 M in methylamine (CH3NH2), a weak base with Kb = 4.4 x 10^-4, and 0.35 M i

n methylammonium bromide (CH3NH3Br).
1. Calculate the pH of the solution. Round your answer to 2 decimal places.
Chemistry
1 answer:
nlexa [21]3 years ago
3 0

Answer:

pH of the solution is 10.37

Explanation:

pOH=pkb+log\frac{[salt]}{[base]}

kb = 4.4 \times 10^{-4}

pkb = -log kb

       = -log4.4 \times 10^{-4}

       = 3.35

salt is methylammonium bromide and methylamine is base

Substitute the values in the above expression as follows:

pOH=pkb+log\frac{[salt]}{[base]} \\=3.35+log\frac{0.35}{0.18} \\=3.35+0.28\\=3.63

pH = 14 - pOH

       = 14 - 3.63

       = 10.37

pH of the solution is 10.37

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