Answer to Question #97180 in General Chemistry for Boye Zephaniah

Question

A weak acid has a ka of 6.2×10^-4 (a)what are the PH and percent ionization of the in 3×10^-3M solution (b) what is the PH and percent ionization of the acid when the solution in (a) is diluted 1000- fold?

Answer 1

Q97180

Solution: Let HA be a weak acid with ionization constant $\alpha$ .

$HA \leftrightarrow H^+ + A^-$ ---(reaction at equilibrium)

$c$ ---(concentrations at $t=0$)

$c(1-\alpha) c\alpha$ $c\alpha$ ---(concentrations at $t=t$ )

Thus, $K_a=[H^+][A^-]/[HA]=6.2*10^{-4}$ ----(given)

$\implies (c\alpha*c\alpha)/c(1-\alpha)=6.2*10^{-4}$

$\therefore c\alpha^2/(1-\alpha)=6.2*10^{-4}$ ----(1)

(a)

Given : $c=3*10^{-3}M$

Using (1), we get;

$(3*10^{-3})\alpha^2+(6.2*10^{-4})\alpha-6.2*10^{-4}=0$

Solving the above quadratic equation for $\alpha$ , we get;

$\alpha=0.363$ (Answer)

(negative root is neglected as percent ionization is always positive)

Thus, percent ionization is $36.3$ %

Now, $pH =(-log_{10}[H^+])=(-log_{10}(c\alpha))$

$=-log_{10}(3*10^{-3}*0.363)$

$=2.963$ (Answer)

pH of the solution at a concentration of $3*10^{-3}M$ is $2.963$

(b)

Now; solution is diluted 1000 times.

Thus, the concentration becomes; $c=3*10^{-6}M$

Again, using (1), we get;

$(3*10^{-6})\alpha^2+(6.2*10^{-4})\alpha-6.2*10^{-4}=0$

Solving the above quadratic equation for $\alpha$ we get;

$\alpha=0.995$ (Answer)

(negative root is neglected as percent ionization is always positive)

Thus, percent ionization is $99.5$ %

Again, $pH =(-log_{10}[H^+])=(-log_{10}(c\alpha))$

$=-log_{10}(3*10^{-6}*0.995)$

$=5.525$ (Answer)

pH of the solution at a concentration of $3*10^{-6}M$ is $5.525$