Question

Calculate the [H+] and pH of a 0.0040 M hydrazoic acid solution. Keep in mind that the Ka of hydrazoic acid is 2.20×10−5. Use the method of successive approximations in your calculations or the quadratic formula.

Answer 1

Answer:

$[H^+]=0.000285$

$pH=3.55$

Explanation:

In this, we can with the ionization equation for the hydrazoic acid ($HN_3$). So:

$HN_3~~H^+~+~N_3^-$

Now, due to the Ka constant value, we have to use the whole equilibrium because this is not a strong acid. So, we have to write the Ka expression:

$Ka=\frac{[H^+][N_3^-]}{[HN_3]}$

For each mol of $H^+$ produced we will have 1 mol of $N_3^-$. So, we can use "X" for the unknown values and replace in the Ka equation:

$Ka=\frac{X*X}{[HN_3]}$

Additionally, we have to keep in mind that $HN_3$ is a reagent, this means that we will be consumed. We dont know how much acid would be consumed but we can express a subtraction from the initial value, so:

$Ka=\frac{X*X}{0.004-X}$

Finally, we can put the ka value and solve for "X":

$2.2X10^-^5=\frac{X*X}{0.004-X}$

$2.2X10^-^5=\frac{X^2}{0.004-X}$

$X= 0.000285$

So, we have a concentration of 0.000285 for $H^+$. With this in mind, we can calculate the pH value:

$pH=-Log[H^+]=-Log[0.000285]=3.55$

I hope it helps!