# What is the pH of a "0.065 M" aqueous solution of acetic acid? K_a = 1.8 xx 10^(-5), and acetic acid, "CH"_3"COOH", reacts with water as shown below: "CH"_3"COOH"(aq) + "H"_2"O"(l) rightleftharpoons "CH"_3"COO"^(-)(aq) + "H"_3"O"^(+)(aq)

Mar 19, 2018

$p H = 2.98$

#### Explanation:

As with all these problems, we first write out the equilibrium equation to inform our reasoning...

HOAc(aq) + H_2O(l) rightleftharpoons ""^(-)OAc + H_3O^+

Now K_"eq"=([H_3O^+][""^(-)OAc])/([HOAc(aq)])

Now INITIALLY, $\left[H O A c\right] = 0.065 \cdot m o l \cdot {L}^{-} 1$..and we PROPOSE that $x \cdot m o l \cdot {L}^{-} 1$ of the acid dissociates....

And so we rewrite the equilibrium expression...

${K}_{\text{eq}} = 1.74 \times {10}^{-} 5 = {x}^{2} / \left(0.065 - x\right)$

...and thus $x = \sqrt{1.74 \times {10}^{-} 5 \times \left(0.065 - x\right)}$

We ASSUME that $0.065 \text{>>} x$..and so...

$x \cong \sqrt{1.74 \times {10}^{-} 5 \times 0.065}$

${x}_{1} = 1.06 \times {10}^{-} 3$..and now we gots an approximation for $x$, we can resubstitute this value back into the equilibrium expression, and see how $x$ evolves....

${x}_{2} = \sqrt{1.74 \times {10}^{-} 5 \times \left(0.065 - 1.06 \times {10}^{-} 3\right)} = 1.05 \times {10}^{-} 3$

${x}_{3} = 1.05 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$...

Since the approximations have converged I am prepared to accept this value...this method of successive approximations is usually more efficient than pfaffing about with the quadratic equation...

And since $x = \left[{H}_{3} {O}^{+}\right] = 1.05 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$...

pH=-log_10[H_3O^+]=-log_10(1.05xx10^-3)=-(-2.98)=??