Question

How many grams of glycine amide should be added to 1.00 g of glycine amide hydrochloride to give 100 mL of solution with pH 8.00

Answer 1

Answer:

0.4235 g of glycine amide should be added

Explanation:

From the given information:

The equation of the reaction can be illustrated as:

$BH^+ \to B^+ + H ^-$

where:

mass of glycine amide hydrochloride = 1.00 g

Suppose x be the amount of (in grams) of glycine amide that is required to be added; Then:

$[BH^+] = \dfrac{1.00 \ g}{110.54 \ gmol \times 0.100 L}$

$[BH^+]=$ 0.0905 M

So, For [B} i.e, for glycine amide

$[B] = \dfrac{x \ g}{74.083 \ gmol \times 0.100 L}$

[B] = 0.135x  M

∴

$\dfrac{[B]}{[BH^+]}=\dfrac{0.135x \ M}{0.0905 \ M }}$

$\dfrac{[B]}{[BH^+]}=1.49x$

The log of the above question can be computed as:

$log _{10}\dfrac{[B]}{[BH^+]}=log _{10}1.49x$

$pH = pK_a + log _{10} \dfrac{[B]}{[BH^+]}$

8.00 = 8.20 + $log _{10}(1.49x)$

8.00 - 8.20 = $log _{10}(1.49x)$

-0.20 = $log _{10}(1.49x)$

1.49 x = $10^{-0.20}$

1.49 x = 0.631

x = 0.631/1.49

x =  0.4235 g