# Solve for c. Ka = (0.02c)(0.02c)/(c-0.02) 1.7x10^-5=(0.02c)(0.02c)/(c-0.02)

## Question:

Solve for c.

{eq}Ka = \frac{(0.02c)(0.02c)}{(c-0.02) 1.7x10^{-5}} = \frac{(0.02c)(0.02c)}{(c-0.02)} {/eq}

## Acid Dissociation:

The acid dissociation constant is an indication of the degree to which the proton on an acid will dissociate (separate) and form a solvated ion in solution. It is related to the concentration of the acid.

To solve for c, cross multiply the fraction:

{eq}1.7 \times 10^{-5} = \frac{\left(0.02c\right) \left(0.02c\right)}{\left(c+0.02\right)} {/eq}.

{eq}1.7 \times 10^{-5}\left(c+0.02\right) - = \left(0.02c\right) \left(0.02c\right) {/eq}.

Expand the brackets

{eq}1.7 \times 10^{-5}c + 3.4 \times 10^{-7} = 0.0004c^2 {/eq}.

Move everything to one side to identify as a quadratic equation (of the form {eq}a^2x + bx + c = 0 {/eq}, where a, b, and c are thecoefficients):

{eq}0 = 0.0004c^2 - 1.7 \times 10^{-5}c - 3.4 \times 10^{-7} {/eq}.

The quadratic formula can be used to solve, whereby only one solution yields a positive, real number:

{eq}\begin{align} x &= \frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ &=\frac{-\left(-1.7 \times 10^{-5}\right) \pm \sqrt{\left(-1.7 \times 10^{-5}\right)^2- 4 \left(0.0004\right) \left(-3.4 \times 10^{-7}\right)}}{2\left(0.0004\right)}\\ &=\boxed{0.06} \end{align} {/eq}