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

0.4235 g of glycine amide should be added

Step-by-step 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^+] = (1.00 \ g)/(110.54 \ gmol * 0.100 L)`

`[BH^+]=` 0.0905 M

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

`[B] = (x \ g)/(74.083 \ gmol * 0.100 L)`

[B] = 0.135x M

∴

`([B])/([BH^+])=(0.135x \ M)/(0.0905 \ M )}`

`([B])/([BH^+])=1.49x`

The log of the above question can be computed as:

`log _(10)([B])/([BH^+])=log _(10)1.49x`

`pH = pK_a + log _(10) ([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