Question

A sample of 50.0 ml of 0.10m nh3(kb=1.8x10-5) is titrated with 0.10m hcl calculate the ph at the equivalence point.

Answer 1

Answer:- pH = 5.28

Solution:- At equivalence point the moles of acid are equivalent to the moles of base.

Ammonia and HCl react in 1:1 mol ratio as is clear from the below balanced equation:

$NH_3(aq)+HCl(aq)\rightarrow NH_4Cl(aq)$

Since the molarity is same for both of them, the volume of HCl required will be same as the given volume of ammonia that is 50.0 mL.

So, total volume of the solution = 50.0 mL + 50.0 mL = 100.0 mL

Convert mL to L as:

$100.0mL(\frac{1L}{1000mL})$ = 0.1L

moles of salt (ammonium chloride) formed will be same equal to the moles of ammonia used. Moles are calculated as:

$50.0mL(\frac{1L}{1000mL})(\frac{0.10mol}{1L})$

= 0.005 mol

So, molarity of ammonium chloride = $\frac{0.005mol}{0.1L}$

= 0.05 M

ammonium chloride is a salt of weak base and strong acid, so the solution will be acidic in nature at equivalence point. The equation could be written below and make the ice table to calculate the pH.

  $NH_4^+(aq)+H_2O(l)\leftrightarrow H_3O^+(aq)+NH_3(aq)$

I                0.05                                                 0                      0

C                -x                                                     +x                   +x

E              0.05-x                                                 x                     x

where x is the change in concentration. Since we have hydronium ion on the product side, Ka is required that we get from given Kb.

$Ka=\frac{H_3O^+][NH_3]}{[NH_4^+]}$

$Ka=\frac{Kw}{kb}$

Where Kw is $1.0*10^-1^4$ .

$Ka=\frac{1.0*10^-^1^4}{1.8*10^-^5}$

$Ka=5.56*10^-^1^0$

let's use this value of Ka to find out the value of x.

$5.56*10^-^1^0=\frac{x^2}{0.05-x}$

x on the bottom could be neglected as the value of Ka is very low.

$5.56*10^-^1^0=\frac{x^2}{0.05}$

On cross multiply and taking square root:

$x=5.27*10^-^6$

From ice table, $[H_3O^+]=x$

So, $[H_3O^+]=5.27*10^-^6$

$pH=-log[H_3O^+]$

$pH=-log(5.27*10^-^6)$

pH = 5.28

Hence, the pH at the equivalence point is 5.28 that clearly indicates the solution is acidic at this point.