Answer:
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
Explanation:
a)
Before the titrant is added; the value for pH of the methylamine is calculated as:
[tex]pOH = \frac{1}{2}[pK_b \ - \ log \ C][/tex]
where ;
[tex]K_b = 4.4 *10^{-4}[/tex]
[tex]pK_b = - log K_b \\ \\ pK_b = log (4.4*10^{-4}) \\ \\ pK_b = 3.36[/tex]
[tex]pOH = \frac{1}{2}[pK_b \ - \ log \ C][/tex]
[tex]pOH = \frac{1}{2}[3.36\ - \ log \0.15][/tex]
[tex]pOH =2.09[/tex]
[tex]pH = 14- pOH \\ \\ pH = 14 - 2.09 \\ \\ pH = 11.91[/tex]
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] = [tex]\frac{3.75}{61.58}[/tex]
[salt] = 0.061 M
[tex]pOH = \frac{1}{2}[pK_w+pK_b+log \ C][/tex]
[tex]pOH = \frac{1}{2}[14+3.36+log \ 0.061][/tex]
[tex]pOH = \frac{1}{2}[16.15][/tex]
[tex]pOH =8.075[/tex]
pOH ≅ 8.08
pH = 14 - 8.08
pH = 5.92
