Kcl Percentage Cb5213

1. **State the problem:** We have a mixture of $KCl$ and $NaBr$ weighing $0.3172$ g. It is titrated with $AgNO_3$ solution to the $Ag_2CrO_4$ endpoint. We want to find the percentage by weight (%w/w) of $KCl$ in the sample. 2. **Relevant chemistry and formula:** The Mohr method uses $AgNO_3$ to precipitate halides as $AgCl$ and $AgBr$. The titration volume corresponds to the total moles of halide ions. The reactions are: $$Ag^+ + Cl^- \rightarrow AgCl(s)$$ $$Ag^+ + Br^- \rightarrow AgBr(s)$$ The total moles of $Ag^+$ used (corrected for blank) equals the total moles of halide ions. 3. **Calculate corrected volume of $AgNO_3$ titrant:** $$V_{corrected} = 36.85\,mL - 0.71\,mL = 36.14\,mL = 0.03614\,L$$ 4. **Calculate moles of $AgNO_3$ (and thus halides):** $$n_{AgNO_3} = M \times V = 0.1120\,mol/L \times 0.03614\,L = 0.00404768\,mol$$ 5. **Define variables:** Let $x$ = moles of $Cl^-$ (from $KCl$) Let $y$ = moles of $Br^-$ (from $NaBr$) Total moles of halides: $$x + y = 0.00404768$$ 6. **Express mass of sample in terms of $x$ and $y$:** Mass of $KCl$: $$m_{KCl} = x \times M_{KCl} = x \times 74.55\,g/mol$$ Mass of $NaBr$: $$m_{NaBr} = y \times M_{NaBr} = y \times 102.89\,g/mol$$ Total mass: $$m_{KCl} + m_{NaBr} = 0.3172\,g$$ 7. **Substitute $y = 0.00404768 - x$ into mass equation:** $$74.55x + 102.89(0.00404768 - x) = 0.3172$$ 8. **Simplify and solve for $x$:** $$74.55x + 0.4165 - 102.89x = 0.3172$$ $$-28.34x = 0.3172 - 0.4165$$ $$-28.34x = -0.0993$$ $$x = \frac{-0.0993}{-28.34} = 0.003504\,mol$$ 9. **Calculate mass of $KCl$:** $$m_{KCl} = 0.003504\,mol \times 74.55\,g/mol = 0.2611\,g$$ 10. **Calculate %w/w $KCl$ in sample:** $$\%w/w\,KCl = \frac{0.2611}{0.3172} \times 100 = 82.3\%$$ **Final answer:** The sample contains approximately $82.3\%$ by weight of $KCl$.