# How do you calculate the pH of 0.15 M aqueous solution of hydrazine?

Aug 11, 2017

We interrogate the reaction.........
${H}_{2} N - N {H}_{2} \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {N}^{+} - N {H}_{2} + H {O}^{-}$ and finally get $p H = 10.6 \ldots \ldots . .$

#### Explanation:

...............

${H}_{2} N - N {H}_{2} \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{3} {N}^{+} - N {H}_{2} + H {O}^{-}$

${K}_{\text{eq}} = \frac{\left[{N}_{2} {H}_{5}^{+}\right] \left[H {O}^{-}\right]}{\left[{N}_{2} {H}_{4}\right]} = 1.0 \times {10}^{-} 6$, and if $x \cdot m o l$ of hydrazine associate, we can put in the numbers........

${K}_{\text{eq}} = 1.0 \times {10}^{-} 6 = \frac{\left(x\right) \times \left(x\right)}{0.15 - x} = {x}^{2} / \left(0.15 - x\right)$

This is a quadratic in $x$ that we could solve exactly. However, because ${K}_{\text{eq}}$ is so small, we are justified in making the approx. $0.15 \text{>>} x$, and we write.........

$1.0 \times {10}^{-} 6 \cong {x}^{2} / 0.15$

${x}_{1} = \sqrt{1.0 \times {10}^{-} 6 \times 0.15} = 3.87 \times {10}^{-} 4$, this is indeed small compared to $0.15 \cdot m o l \cdot {L}^{-} 1$, but now we have an approximation for $x$ we can plug this value back into the equation to get ${x}^{2}$...

${x}_{2} = 3.87 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1$, and given the convergence of ${x}_{1}$ and ${x}_{2}$ we are satisfied that these values are as good as if we used the quadratic equation............

But we are not finished there.....we know that ..............

$\left[{N}_{2} {H}_{5}^{+}\right] = \left[H {O}^{-}\right] = 3.87 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1$...

We DO NOT KNOW $p H$.

But in aqueous solution, $14 = p H + p O H$

$p O H = - {\log}_{10} \left[H {O}^{-}\right] = - {\log}_{10} \left(3.87 \times {10}^{-} 4\right) = 3.41$

$p H = 14 - 3.41 = 10.6$