Question

Consider the titration of $\mathbf{50}\mathbf{.}\mathbf{00}\mathbf{(}\mathbf{\pm }\mathbf{005}\mathbf{)}\mathbf{mL}$of a mixture of ${\mathbf{l}}^{\mathbf{-}}$and ${\mathbf{SCN}}^{\mathbf{-}}$with $\mathbf{0}\mathbf{.}\mathbf{0683}\mathbf{(}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{0001}\mathbf{)}{\mathbf{MAg}}^{\mathbf{+}}$.The first equivalence point is observed at $\mathbf{12}\mathbf{.}\mathbf{6}\mathbf{(}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{)}\mathbf{mL}\mathbf{,}$and the second occurs at $\mathbf{27}\mathbf{.}\mathbf{7}\mathbf{(}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{)}\mathbf{mL}\mathbf{.}$

(b) Suppose that the uncertainties are all the same, except that the uncertainty of the first equivalence point $\mathbf{(}\mathbf{12}\mathbf{.}\mathbf{6}\mathbf{\pm }\mathbf{?}\mathbf{}\mathbf{mL}\mathbf{)}$is variable. What is the maximum uncertainty (milliliters) of the first equivalence point if the uncertainty in ${\mathbf{SCN}}^{\mathbf{-}}$molarity is to be $\mathbf{\le }\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\%}$?

The excess acid required $\mathbf{39}\mathbf{.}\mathbf{96}\mathbf{}\mathbf{mL}$of $\mathbf{0}\mathbf{.}\mathbf{1004}\mathbf{MNaOH}$For complete titration to a phenolphthalein end point. Find the weight percent of calcite in the limestone.

Short answer

The extreme uncertainty of first equivalence point =$0.5\mathrm{mL}$

Step-by-step solution

Define the equivalence point.

An equivalency point is a word used in the titration process. It technically applies to any acid-base or neutralisation reaction.

In a titration, the equivalency point is reached when the amount of titrant administered is sufficient to totally neutralise the analyte solution.

The moles of titrant (standard solution) are equivalent to the moles of the unknown-concentration solution. Because the moles of acid are equal to the quantity needed to neutralise the equivalent moles of base, this is also known as the stoichiometric point

Calculate extreme uncertainty of first equivalence point

$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{error}\mathrm{in}15.1\mathrm{mL}\mathrm{is}\mathrm{y}\% \\ ({4_2}^{.0\%)}={(\mathrm{y}\%)}^2+{(0.146\%)}^2+{(0.100\%)}^2 \\ y=4.00\%=0.603mL \\ 27.7(\pm 0.3)-12.6(\pm ?)=15.1(\pm 0.603) \\ 27.7(\pm 0.3)-12.6(\pm ?)=15.1(\pm 0.603) \\ =0.5\mathrm{mL} \end{gathered}$$

Thus, the extreme uncertainty of first equivalence point$=0.5\mathrm{mL}$