Question

Using the quadratic equation to calculate [H3O+] in 0.00250 M HNO2, what are the values of a, b, c and x , where a, b, and c are the coefficients in the quadratic equation ax2+bx+c=0, and x is [H3O+]? Recall that Ka=4.5×10−4 .

Express a, b, c, and x numerically separated by commas.

Answer 1

Question #79016

Using the quadratic equation to calculate $\left[\mathrm{H}3\mathrm{O}+\right]$ in 0.00250 M HNO$_2$, what are the values of a, b, c and x, where a, b, and c are the coefficients in the quadratic equation $\alpha x^2 + \beta x + \gamma c = 0$, and x is $\left[\mathrm{H}3\mathrm{O}+\right]$? Recall that $\mathrm{Ka} = 4.5 \times 10^{-4}$.

Express a, b, c, and x numerically separated by commas.

The right answer is $1, 4.5 \times 10^{-4}, -1.125 \times 10^{-6}, 0.0104$.

Solution:

As $\mathrm{HNO}_2$ is a weak acid, the formula for calculation concentration of $\left[\mathrm{H}_3\mathrm{O}^+\right]$ is [1]:

$$K_{\alpha} = \frac{x^2}{C_{HA} - x}$$

where $x$ — the concentration of $\left[\mathrm{H}_3\mathrm{O}^+\right]$.

After transformation, we will get the next equation:

$$x^2 + K_{\alpha} * x - C_{HA} * K_{\alpha} = 0$$

So, $a = 1$, $b = K_a = 4.5 \times 10^{-4}$, $c = -C_{HA} * K_a = -4.5 \times 10^{-4} * 0.0025 = -1.125 \times 10^{-6}$.

The concentration of $\left[\mathrm{H}_3\mathrm{O}^+\right]$ is equal to:

$$x = \frac{-K_{\alpha} + \sqrt{K_{\alpha} + 4 * K_{\alpha} * C_{HA}}}{2} = 0.0104$$

So, $x = 0.0104$.

So, the full answer is $a = 1$, $b = 4.5 \times 10^{-4}$, $c = -1.125 \times 10^{-6}$, $c = 0.0104$.

Reference:

[1] http://www.science.uwaterloo.ca/~cchieh/cact/c123/wkacids.html