Suppose that you used smaller volumetric apparatus to prepare $0.250{\mathrm{MNH}}_3$solution by diluting role="math" localid="1663558223400" $84.6\left(\pm 0.8\right)\mathrm{\mu L}$ of $28.0\left(\pm 0.5\right)$ wt\% ${\mathrm{NH}}_3$ up to $5.00\left(\pm 0.02\right)\mathrm{mL}$. Find the uncertainty in0.250M.(Answer: $0.250\left(\pm 0.005\right)\mathrm{M}$.The uncertainty in density of concentrated${\mathrm{NH}}_3$overwhelms all other smaller uncertainties in this procedure.)

Uncertainty refers to the inability to forecast a value. The last digit in a measured value will have a related uncertainty. Absolute and relative uncertainties are the two types of uncertainty.

Absolute ambiguity:

The marginal value of a measurement is expressed in this way.

Uncertainty in a relative sense:

It compares the extent of the absolute uncertainty to the size of the measurement it is connected with.

$\mathrm{Relative}\mathrm{uncertainty}=\frac{\mathrm{absolute}\mathrm{uncerntainty}}{\mathrm{magnitude}\mathrm{of}\mathrm{measurment}}$.

For a set of measurements having uncertainty values as $e_1,e_2$and $e_3$, the uncertainty $\left(e_3\right)$in addition and subtraction can be calculated as follows:

$e_4=\sqrt{e_1^2+e_2^2+e_3^2}$.

Given information:

A $0.250{\mathrm{MNH}}_3$ solution is prepared by diluting $84.6\left(\pm 0.8\right)\mathrm{\mu L}\mathrm{of}28.0\left(\pm 0.5\right)$wt % ${\mathrm{NH}}_3$up to $5.00\left(\pm 0.02\right)\mathrm{mL}$

Density $=0.899\left(\pm 0.003\right)\mathrm{g}/\mathrm{mL}$.

All other lesser uncertainties are overwhelmed by the uncertainty in the concentrated ${\mathrm{NH}}_3$density.

To compute the uncertainty in the molarity, first calculate the uncertainty in the moles delivered, which is 5.00mL to the flask.

The concentrated reagent has $0.899\left(\pm 0.003\right)\mathrm{g}$of the solution per mL .

From the given weight percentage, localid="1663559051222" $28.0\left(\pm 0.5\right)wt\%$, we know that the reagent contains $28.0\left(\pm 0.005\right)\mathrm{g}$of NH₃ per gram of the solution.

For multiplication, convert the absolute uncertainties into the percent relative uncertainty.

For $0.899\left(\pm 0.003\right)$,

Percent relative uncertainty $=\frac{0.003}{0.899}\times 100$.

$=\pm 0.334\%$

For$0.280\left(\pm 0.005\right)$,

Percent relative uncertainty$=\frac{0.005}{0.280}\times 100$.

$=\pm 0.334\%$

Grams of NH₃ per mL

in the concentrated reagent

$$\begin{gathered} =0.899\left(\pm 0.003\right)\frac{\mathrm{g}\mathrm{solution}}{\mathrm{mL}}\times 0.280\left(\pm 0.005\right)\frac{{\mathrm{gNH}}_3}{\mathrm{g}\mathrm{solution}} \\ =0.899\left(\pm 0.334\%\right)\frac{\mathrm{g}\mathrm{solution}}{\mathrm{mL}}\times 0.280\left(\pm 1.79\%\right)\frac{{\mathrm{gNH}}_3}{\mathrm{g}\mathrm{solution}} \\ =0.2517\left(\pm 1.82\%\right)\frac{\mathrm{gNH}}{\mathrm{mL}} \end{gathered}$$

The moles of ammonia present in a concentrated reagent of $84.6\left(\pm 0.8\right)\mathrm{\mu L}$$ are computed as follows:

Convert the volume in microliter to milliliter as$0.0846\left(\pm 0.0008\right)\mathrm{mL}$.

The relative uncertainty in volume is $0.0008/0.0846=0.946\%$.

$$\begin{gathered} \mathrm{molNH}3=\frac{0.2517\left(\pm 1.82\%\right){\displaystyle \frac{2{\mathrm{NH}}_3}{\mathrm{ml}.}}\times 0.0846\left(\pm 0.946\%\right)\mathrm{mL}}{17.031\left(\pm 0\%\right){\displaystyle \frac{2{\mathrm{NH}}_3}{\mathrm{mal}}}} \\ =0.00125\left(\pm 2.05\right)\mathrm{mol} \end{gathered}$$

As$\sqrt{{\left(1.82\%\right)}^2+{\left(0.946\%\right)}^2+{\left(0\%\right)}^2}=2.05\%$,

this much of ammonia was diluted to$5.00\left(\pm 0.02\right)\mathrm{mL}$.

Convert millilitre to liter as$0.005\left(\pm 0.00002\right)\mathrm{mL}$.

The relative uncertainty in the final volume is $0.00002/0.005=0.004\%$.

The molarity is calculated as follows:

$$\begin{gathered} \mathrm{Molarity}=\frac{\mathrm{moles}}{\mathrm{liter}} \\ =\frac{0.00125\left(\pm 2.05\%\right)\mathrm{mL}}{0.005\left(\pm 0.004\%\right)\mathrm{L}} \\ =0.250\left(\pm 2.05\%\right)\mathrm{M} \end{gathered}$$

As$\sqrt{{\left(2.05\%\right)}^2+{\left(0.004\%\right)}^2}=2.05\%$,

The absolute uncertainty is $2.05\%$of $0.250\mathrm{M}=0.005$.

Therefore, the uncertainty in the concentration of ammonia $\left[{\mathrm{NH}}_3\right]$is $0.250\left(\pm 0.005\right)M$.