Question

Determine ph buffer containing 120.5 g mathylamine and 230.5 g ch3nh3 f in 1 litre water

Answer 1

According to Hendersen-Hasselbalch equation,

$pH=pK_{a}+log\frac{[Base]}{[Acid]}$

$pK_{b}$ of methyl amine = 3.36

$pK_{a}$ = 14.00-p$K_{b}$

= 14.00 - 3.36

= 10.64

$[CH_{3}NH_{2}] = 120.5 g CH_{3}NH_{2} (\frac{1 molCH_{3}NH_{2}}{31.06gCH_{3}NH_{2}}) * \frac{1}{1 L}$

= 3.88 M

$[CH_{3}NH_{3}^{+}] =230.5 g CH_{3}NH_{3}^{+} (\frac{1 molCH_{3}NH_{3}^{+} }{32.06 g CH_{3}NH_{3}^{+}}) *\frac{1}{1 L}$

= 7.19 M

$pH = 10.64 + log\frac{3.88 M}{7.19 M}$

pH = 10.37