Question

Calculate the ph at the equivalence point for the titration of 0.230 m methylamine (ch3nh2) with 0.230 m hcl. the kb of methylamine is 5.0× 10–4.

Answer 1

Answer: The pH at the equivalence point for the titration will be 0.65.

Solution:

Let the concentration of $[OH^-]$ be x

Initial concentration of $[CH3NH_2]$, c = 0.230 M

        $CH_3NH_2+H_2O\rightleftharpoons CH_3NH_3^++OH^-$

at eq'm  c-x                         x                    x

Expression of $K_b$:

$K_b=\frac{[CH_3NH_3^+][+OH^-]}{[CH_3NH_2]}=\frac{x\times x}{c-x}=\frac{x^2}{c-x}$

Since ,methyl-amine is a weak base,c>>x so $c-x\approx c$.

$K_b=\frac{x^2}{c}=5.0\times 10^{-4}=\frac{x^2}{0.230 M}$

Solving for x, we get:

$x=1.07\times 10^{-2} M$

Given, HCl with 0.230 M , it dissociates fully in water which means $[H^+]$ = 0.230 M

$[OH^-]=[H^+]$ will result in neutral solution, since $[OH^-]<[H^+]$

Remaining $[H^+]$ after neutralizing $[OH^-]$ions

$[H^+]_{\text{left in solution}}=[H^+]-[OH^-]=0.230-1.07\times 10^{-2}=0.2193 M$

$pH=-log{[H^+]_{\text{left in solution}}=-log(0.2193)=0.65$

The pH at the equivalence point for the titration will be 0.65.

Answer 2

0.65 is the pH at the equivalence point for the titration of 0.230 m methylamine with 0.230 m HCl.

How we calculate pH of any solution?

pH of any solution will be calculated as the negative log of the concentration of H⁺ ions and it is represented as:

pH = -log[H⁺]

Given reaction with ICE table will be represented as:

                         CH₃NH₂ + H₂O ⇄ CH₃NH₃⁺ + OH⁻

Initial:                  0.230                         0             0

Change:                -x                            +x            +x

Equilibrium:      0.230-x                      +x            +x

Equation for Kb will be written as:

Kb = [CH₃NH₃⁺][OH⁻] / [CH₃NH₂]

Kb = x² / 0.230-x

Given value of Kb = 5.0× 10⁻⁴

Given base is weak base and it shows weak dissociation, so we ignore the value of x for CH₃NH₂ and equation becomes-

5.0 × 10⁻⁴ = x² / 0.230

x = 1.07 × 10⁻²M

Here HCl is a strong acid and it completely dissociates into ions, so the concentration of H⁺ ions is also 0.230 M.

Here the concentration of H⁺ ions is in excess and their values of excess H⁺ ions are 1.07 × 10⁻² - 0.230 = 0.2193 M

Now, pH = -log(0.2193) = 0.65

Hence, the pH is 0.65.

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