A solution of $\ce{F-}$ is prepared by dissolving $(0.0933\pm0.0004)\ \mathrm g$ of $\ce{NaF}$ $(M= (41.989\pm0.001)\ \mathrm {g/mol})$ in $156.00\pm\ \mathrm{mL}$ of water. Calculate the concentration of $\ce{F-}$ in the solution and its absolute uncertainty.
First of all, uncertainty aside, I got $0.0141$ for the concentration of $\ce{F-}$. However, this is not the correct answer (according to Sapling Learning).
Furthermore, what technique should I use to find the absolute uncertainty?
Generally, it is preferable to write down the complete quantity equation before plugging in the numbers. If necessary, perform all mathematic operations on quantities symbolically. Do not plug in the numbers until you have only one equation for the desired result. The same applies to the equation for the uncertainty of the desired result.
In this case, the concentration $c$ shall be calculated. The mass $m=0.0933\ \mathrm g$, the molar mass $M=41.989\ \mathrm{g\ mol^{-1}}$, and the volume $V=156.00\ \mathrm{ml}$ are given. The uncertainty of the mass $m$ and the molar mass $M$ are given as $u(m)=0.0004\ \mathrm g$ and $u(M)=0.001\ \mathrm{g\ mol^{-1}}$, respectively. The uncertainty of the volume $V$ is not given; however, when a number is given without any further information, it is generally interpreted so that the last digit is rounded. Therefore, the given value $V=156.00\ \mathrm{ml}$ is assumed to represent a value between $155.995\ \mathrm{ml}$ and $156.005\ \mathrm{ml}$, or $V=\left(156.000\pm0.005\right)\ \mathrm{ml}$. Thus, the assumed uncertainty of $V$ is $u(V)=0.005\ \mathrm{ml}$.
The concentration $c$ is defined as $$c=\frac nV\tag1$$ where $n$ is amount of substance, which is unknown. Since the molar mass $M$ is defined as $$M=\frac mn\tag2$$ the concentration $c$ can be calculated according to $$\begin{align}c&=\frac m{M\cdot V}\tag3\\[6pt] &=\frac {0.0933\ \mathrm g}{41.989\ \mathrm{g\ mol^{-1}}\times 156.00\ \mathrm{ml}}\\[6pt] &=1.42437\times10^{-5}\ \mathrm{mol\ ml^{-1}}\\[6pt] &=0.0142437\ \mathrm{mol\ l^{-1}} \end{align}$$
In this case, we may assume that the uncertainties of $m$, $M$ and $V$ are not correlated. Therefore, the relative uncertainty of the concentration $c$ may be estimated as $$\begin{align} \left(\frac{u(c)}{c}\right)^2 &= \left(\frac{u(m)}{m}\right)^2 + \left(\frac{u(M)}{M}\right)^2 + \left(\frac{u(V)}{V}\right)^2\tag4 \\[6pt] \frac{u(c)}{c} &= \sqrt{\left(\frac{u(m)}{m}\right)^2 + \left(\frac{u(M)}{M}\right)^2 + \left(\frac{u(V)}{V}\right)^2}\tag5 \end{align}$$
Thus, the absolute uncertainty of the concentration $c$ is
$$\begin{align} u(c) &= \sqrt{\left(\frac{u(m)}{m}\right)^2 + \left(\frac{u(M)}{M}\right)^2 + \left(\frac{u(V)}{V}\right)^2} \cdot c\tag6 \\[6pt] &= \sqrt{\left(\frac{u(m)}{m}\right)^2 + \left(\frac{u(M)}{M}\right)^2 + \left(\frac{u(V)}{V}\right)^2} \cdot \frac m{M\cdot V}\tag7 \\[6pt] &= \sqrt{\left(\frac{0.0004\ \mathrm g}{0.0933\ \mathrm g}\right)^2 + \left(\frac{0.001\ \mathrm{g\ mol^{-1}}}{41.989\ \mathrm{g\ mol^{-1}}}\right)^2 + \left(\frac{0.005\ \mathrm{ml}}{156.00\ \mathrm{ml}}\right)^2} \cdot \frac {0.0933\ \mathrm g}{41.989\ \mathrm{g\ mol^{-1}}\times 156.00\ \mathrm{ml}} \\[6pt] &=6.10687\times10^{-8}\ \mathrm{mol\ ml^{-1}}\\[6pt] &\approx0.00006\ \mathrm{mol\ l^{-1}} \end{align}$$
Note that the same result may be obtained using the general formula
$$\begin{align} u(c) &= \sqrt{\left(\frac{\partial c}{\partial m}\right)^2\cdot u^2(m)+\left(\frac{\partial c}{\partial M}\right)^2\cdot u^2(M)+\left(\frac{\partial c}{\partial V}\right)^2\cdot u^2(V)}\tag8 \\[6pt] &= \sqrt{\left(\frac1{M\cdot V}\right)^2\cdot u^2(m)+\left(-\frac m{M^2\cdot V}\right)^2\cdot u^2(M)+\left(-\frac m{M\cdot V^2}\right)^2\cdot u^2(V)}\tag9 \\[6pt] &= \sqrt{\left(\frac1{41.989\ \mathrm{g\ mol^{-1}}\times 156.00\ \mathrm{ml}}\right)^2\times \left(0.0004\ \mathrm g\right)^2\\ +\left(-\frac {0.0933\ \mathrm g}{\left(41.989\ \mathrm{g\ mol^{-1}}\right)^2\times 156.00\ \mathrm{ml}}\right)^2\times \left(0.001\ \mathrm{g\ mol^{-1}}\right)^2\\ +\left(-\frac {0.0933\ \mathrm g}{41.989\ \mathrm{g\ mol^{-1}}\times \left(156.00\ \mathrm{ml}\right)^2}\right)^2\times \left(0.005\ \mathrm{ml}\right)^2}\\[6pt] &=6.10687\times10^{-8}\ \mathrm{mol\ ml^{-1}}\\[6pt] &\approx0.00006\ \mathrm{mol\ l^{-1}} \end{align}$$
The numerical values of the concentration $c$ and its uncertainty $u(c)$ should not be given with an excessive number of digits. In this case, it suffices to quote $u(c)=0.00006\ \mathrm{mol\ l^{-1}}$ and accordingly $c=0.01424\ \mathrm{mol\ l^{-1}}$; i.e. $c=\left(0.01424\pm0.00006\right)\ \mathrm{mol\ l^{-1}}$ or $c=0.01424(6)\ \mathrm{mol\ l^{-1}}$. In some cases, however, it may be necessary to retain additional digits to avoid round-off errors in subsequent calculations.
Note
The large equations for the uncertainty $u(c)$ may be confusing and susceptible to errors.
To verify the result, you should check whether the calculated quantity and its absolute uncertainty have the same unit (here: $\left[c\right]=\left[u(c)\right]=\mathrm{mol\ ml^{-1}}$).
Furthermore, you should compare the relative uncertainty of the result with the relative uncertainty of the input data. Usually, the relative uncertainty of the result is greater than or equal to the relative uncertainty of the input data. In this case:
$$\frac{u(m)}{m}=\frac{0.0004\ \mathrm g}{0.0933\ \mathrm g}=0.0042872\approx0.43\ \%$$
$$\frac{u(M)}{M}=\frac{0.001\ \mathrm{g\ mol^{-1}}}{41.989\ \mathrm{g\ mol^{-1}}}=2.38158\times10^{-5}\approx0.0024\ \%$$
$$\frac{u(V)}{V}=\frac{0.005\ \mathrm{ml}}{156.00\ \mathrm{ml}}=3.20513\times10^{-5}\approx0.0032\ \%$$
$$\frac{u(c)}{c}=\frac{6.10687\times10^{-8}\ \mathrm{mol\ ml^{-1}}}{1.42437\times10^{-5}\ \mathrm{mol\ ml^{-1}}}=0.0042874\approx0.43\ \%$$
We find that, in this case, the uncertainty of the concentration $c$ is mainly caused by the uncertainty of the mass $m$.