Barium titanate, a ceramic used in electronics, was analyzed by the following procedure: Into a Pt crucible was placed \(1.2\;{\rm{g}}\)of \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}\) and \(0.8\;{\rm{g}}\)of \({\rm{N}}{{\rm{a}}_2}\;{{\rm{B}}_4}{{\rm{O}}_7}\)plus \(0.3146\;{\rm{g}}\)of unknown. After fusion at \({1000^\circ }{\rm{C}}\)in a furnace for\(30\;{\rm{min}}\), the cooled solid was extracted with \(50\;{\rm{mL}}\)of\(6{\rm{MHCl}}\), transferred to a \(100 - {\rm{mL}}\) volumetric flask, and diluted to the mark. A \(25.00 - {\rm{mL}}\)aliquot was treated with \(5\;{\rm{mL}}\)of \(15\% \)tartaric acid (which complexes \({\rm{T}}{{\rm{i}}^{4 + }}\)and keeps it in aqueous solution) and \(25\;{\rm{mL}}\)of ammonia buffer,\({\rm{pH}}9.5\). The solution was treated with organic reagents that complex\({\rm{B}}{{\rm{a}}^{2 + }}\), and the \({\rm{Ba}}\)complex was extracted into \({\rm{CC}}{{\rm{l}}_4}.\)After acidification (to release the \({\rm{B}}{{\rm{a}}^{2 + }}\) from its organic complex), the \({\rm{B}}{{\rm{a}}^{2 + }}\)was backextracted into\(0.1{\rm{MHCl}}\). The final aqueous sample was treated with ammonia buffer and methylthymol blue (a metal ion indicator) and titrated with \(32.49\;{\rm{mL}}\) of \(0.01144{\rm{M}}\)EDTA. Find the weight per cent of Ba in the ceramic.

Step by step solution

01

Defining the weight percent.

The Weight Percentage is simply the ratio of a solute's mass to the mass of a solution multiplied by 100. The Weight Percentage is also referred to as the Mass Percentage.

\(Percent by weight = \frac{{{\rm{ gram of solute }}}}{{100g{\rm{ of solution }}}}\)

02

Determining the moles of EDTA and \({\rm{B}}{{\rm{a}}^{2 + }}\).

Here \(1/4\)of the sample is required for EDTA titration so we will first calculate the moles of EDTA spent on titration:

\(n({\rm{EDTA}}) = c \times V\)

\(n({\rm{EDTA}}) = 0.01144{\rm{M}} \times (0.03249\;{\rm{L}})\)

\(n({\rm{EDTA}}) = 3.717 \times {10^{ - 4}}\;{\rm{mol}}\)

Next we will determine the moles of \({\rm{B}}{{\rm{a}}^{2 + }}\)and calculate its mass:

\(n\left( {{\rm{B}}{{\rm{a}}^{2 + }}} \right) = 4 \times n({\rm{ EDTA }})\)

\(n\left( {{\rm{B}}{{\rm{a}}^{2 + }}} \right) = 4 \times 3.717 \times {10^{ - 4}}\;{\rm{mol}} = 1.487 \times {10^{ - 3}}\;{\rm{mol}}\)

\( \to m\left( {{\rm{B}}{{\rm{a}}^{2 + }}} \right) = 0.2042\;{\rm{g}}\)

03

Determining the weight percent.

Now calculating the weight percent

\(wt\% = \frac{{m\left( {{\rm{B}}{{\rm{a}}^{2 + }}} \right)}}{{m({\rm{ unknown }})}}\)

\(wt\% = \frac{{0.2042\;{\rm{g}}}}{{0.3146\;{\rm{g}}}}\)

\(wt\% = 0.6049\)

Therefore the weight percent of Ba in the ceramic is \(0.6049\)%.

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