Question

If uncertainty in pHis doubledto$\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{06}$, what is the relative uncertainty in$\left[H^{+}\right]$?

Short answer

An uncertainty of 0.06 (doubled) in pH gives an uncertainty of 14% in$\left[{\mathrm{H}}^{+}\right]$

Step-by-step solution

Exponents and logarithm

Consider, y = antilog,$\mathit{x}\mathbf{\Rightarrow }\mathit{y}\mathbf{=}{\mathbf{10}}^{\mathbf{x}}$

Here, the relative uncertainty in y is proportional to the absolute uncertainty in x.

Uncertainty for role="math" localid="1663315267361" ${10}^x$

role="math" localid="1663315330316" $$\begin{gathered} y={10}^x \\ \frac{e_y}{y}=\left(in10\right)e_x \\ \approx 2.3026e_x \end{gathered}$$

Find the relative uncertainty in [H+]

The given pH is $5.21\pm 0.03$

The uncertainty in pH is doubled to $\pm 0.06$

Let us calculate the uncertainty in $\left[H^{+}\right]$

Now, according to $pH=-\log \left[H^{+}\right]$

The concentration of hydrogen ion $\left[H^{+}\right]$can be calculated by rearranging the above pH equation as $\left[H^{+}\right]={10}^{-pH}$

Insert the value ofin the above equation.

$$\begin{gathered} \left[H^{+}\right]={10}^{-pH} \\ ={10}^{-\left(5.21\pm 0.06\right)} \end{gathered}$$

This is relevant to the equation $y={10}^x$,where $y=\left[H^{+}\right]$and$x=-\left(5.21\pm 0.06\right)$

Thus,

$$\begin{gathered} \frac{e_y}{y}=2.3026 \\ {e}_x\frac{e_{\left[H^{+}\right]}}{\left[H^{+}\right]}=2.3026 \\ {e}_{pH}=2.3026\left(0.06\right) \\ =0.138156 \end{gathered}$$

The relative uncertainty in$\left[H^{+}\right]$ is 0.138156

Calculate the hydrogen ion concentration as shown below.

$$\begin{gathered} \left[H^{+}\right]={10}^{-pH} \\ ={10}^{-5.21} \\ =6.17\times {10}^{-6}M \end{gathered}$$

Let us find$e_{\left[H^{+}\right]}$

$$\begin{gathered} \frac{e_{\left[H^{+}\right]}}{\left[H^{+}\right]}=\frac{e_{\left[H^{+}\right]}}{6.17\times {10}^{-6}M} \\ 0.138156=\frac{e_{\left[H^{+}\right]}}{6.17\times {10}^{-6}M} \\ {e}_{\left[H^{+}\right]}=\left(0.138156\right)\left(6.17\times {10}^{-6}M\right) \\ =0.85242252\times {10}^{-6} \\ =8.5242252\times {10}^{-7} \end{gathered}$$

Therefore,$e_{\left[H^{+}\right]}=8.52\times {10}^{-7}$

Thus, the concentration of $\left[H^{+}\right]$is$6.2\left(\pm 0.9\right)\times {10}^{-6}M$

An uncertainty of 0.03 in pH gives an uncertainty of 7% in $\left[H^{+}\right]$

Similarly, an uncertainty of 0.06 (doubled) in gives an uncertainty of $0.138156\times 100=13.81\%$in $\left[H^{+}\right]$