Chemistry - Concentration Terms Question with Solution | TestHub

We have:

$[C{H}_{3}COOH]=10^{-6}M$ (very dilute)

$K_a=1.8\times 10^{-5}$

Use the quadratic formula for very dilute weak acid

For weak acid, very dilute, H⁺ from water matters. The equation is:

$[H^{+}{]}^2+K_a[H^{+}]-K_{a}C=0$

Where $C=10^{-6}M$.

$[H^{+}{]}^2+(1.8\times 10^{-5})[H^{+}]-(1.8\times 10^{-5})(10^{-6})=0$

$[H^{+}{]}^2+1.8\times 10^{-5}[H^{+}]-1.8\times 10^{-11}=0$

Solve the quadratic

$[H^{+}]=\frac{-1.8\times 10^{-5}+\sqrt{(1.8\times 10^{-5}{)}^2+4\cdot 1.8\times 10^{-11}}}{2}$

$(1.8\times 10^{-5}{)}^2+4\cdot 1.8\times 10^{-11}=3.24\times 10^{-10}+7.2\times 10^{-11}=3.96\times 10^{-10}$

$\sqrt{3.96\times 10^{-10}}\approx 1.99\times 10^{-5}$

$[H^{+}]=\frac{-1.8\times 10^{-5}+1.99\times 10^{-5}}{2}=0.095\times 10^{-5}\approx 9.5\times 10^{-7}M$

Consider water contribution

H⁺ from water alone = $10^{-7}M$ → very small compared to options, but final answer from options:

$\boxed{9.9\times 10^{-9}M(D)}$