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Hatshy [7]
3 years ago
6

Calculate the pH at the equivalence point for the titration of 0.190 M 0.190 M methylamine ( CH 3 NH 2 ) (CH3NH2) with 0.190 M H

Cl . 0.190 M HCl. The K b Kb of methylamine is 5.0 × 10 − 4 .
Chemistry
1 answer:
WARRIOR [948]3 years ago
6 0

Answer:

5.86

Explanation:

The equation of reaction is between methylamine and hydrochloric acid

CH₃NH₂   +   HCl -------> CH₃NH⁺₃     +      Cl⁻

At equivalence point, there is a mixture of equal volume of acid and base. As such, the concentration of CH₃NH⁺₃ will be half the initial starting concentration of the reactant.

i.e [CH₃NH⁺₃] = \frac{0.190}{2}

= 0.095 M

Now, the dissociation of CH₃NH⁺₃ yields

CH₃NH⁺₃       ⇄      CH₃NH₂      +      H⁺

The ICE Table can be constructed as follows:

                      CH₃NH⁺₃       ⇄      CH₃NH₂      +      H⁺

Initial                0.095                        0                      0

Change            -x                                x                      x

Equilibrium    (0.095 - x)                     x                     x

K_a =\frac{[CH_3NH_2][H^+]}{[CH_3NH^+_3]}

K_a =  \frac{[x][x]}{[0.095-x]}

K_a =  \frac{[x^2]}{[0.095-x]}     ------ equation (1)

We all know that:

K_a = \frac{K_w}{K_b}

and we are given K_b = 5.0*10^{-4;  and K_w = 10^{-14 ;

Then;

K_a= \frac{10^{-14}}{5*10^{-4}}

K_a = 2*10^{-11}

We can now re-write the equation (1) to be :

2*10^{-11} = \frac{x^2}{0.095-x}

x^2 = 2*10^{-11} (0.095 -x)

x^2 = 1.9*10^{-12} - 2*10^{-11}x

x^2 + (2*10^{-11})x - (1.9*10^{-12})  = 0

where a = 1; b = 2×10⁻¹¹; c = 1.9×10⁻¹²

Using the quadratic formula; \frac{-b+/-\sqrt{b^2-4ac} }{2a} ; we have

= \frac{-(2*10^{-11})+\sqrt{(-(2*10^{-11})^2-4(1)(1.9*10^{-12})} }{2*1}  OR \frac{-(2*10^{-11})-\sqrt{(-(2*10^{-11})^2-4(1)(1.9*10^{-12})} }{2*1}

x = 1.38*10^{-6}M

[CH₃NH₂] = [H⁺] =  1.38*10^{-6}M

∴ pH = -log [H⁺]

pH = -log 1.38*10^{-6}M

pH = 5.86

Hence, the  pH at the equivalence point for the titration  = 5.86

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