Question

A 25.00-mL solution of 0.1500 M methylamine (CH3NH2) is titrated with a standardized 0.1025 M solution of HCl at 25°C. Enter your numbers to 2 decimal places. Kb = 4.4x10-4 What is the pH of the methylamine solution before titrant is added? 11.91 How many milliliters of titrant are required to reach the equivalence point? 36.59 What is the pH at the equivalence point?

Answer 1

a) pH of methylamine = 11.91

b) Volume of milliliters of titrant required to reach equivalence point = 36.59 mL

c) The pH at equivalence point = 5.92

Step-by-step explanation:

a)

Before the titrant is added; the value for pH of the methylamine is calculated as:

`pOH = (1)/(2)[pK_b \ - \ log \ C]`

where ;

`K_b = 4.4 *10^(-4)`

`pK_b = - log K_b \\ \\ pK_b = log (4.4*10^(-4)) \\ \\ pK_b = 3.36`

`pOH = (1)/(2)[pK_b \ - \ log \ C]`

`pOH = (1)/(2)[3.36\ - \ log \0.15]`

`pOH =2.09`

`pH = 14- pOH \\ \\ pH = 14 - 2.09 \\ \\ pH = 11.91`

b)

How many milliliters of titrant are required to reach the equivalence point?

Millimoles of base = (25.00 mL × 0.1500 M) of methylamine = 3.75

3.75 millimoles of HCl is required to reach equivalence point.

3.75 = Volume × 0.1025

Volume of milliliters of titrant required to reach equivalence point = 36.59 mL

c)

The total volume = 36.58 + 25.00 = 61.58 mL

Concentration of the salt; i.e [salt] =
`(3.75)/(61.58)`

[salt] = 0.061 M

`pOH = (1)/(2)[pK_w+pK_b+log \ C]`

`pOH = (1)/(2)[14+3.36+log \ 0.061]`

`pOH = (1)/(2)[16.15]`

`pOH =8.075`

pOH ≅ 8.08

pH = 14 - 8.08

pH = 5.92