No 17 1. How many milliliters of 0.10MH_(2)SO_(4) must be added to 100 mL of 0.10MNaOH to give a solution that is 0.050 M in H_(2)SO_(4) ? Assume volumes are additive. 2. A sample of pure KHC_(2)O_(4)cdot H_(2)C_(2)O_(4)cdot 2H_(2)O (three replaceabl e hydrogens)requires 46.2mL of 0.100 M NaOH for titration.How many milliliters M KMnO_(4) will the same-size sample react? 3. A sample containing ammonium chloride was warmed with 150.0 mL of 1.00mol/L sodium hydroxide.After all the ammonia had been driven off, the excess sodium hydroxide required 50.00 mL of 0.150mol/L sulfuric acid for neutralizatior 1. What mass of ammonium chloride did the sample contain? 4. FeCl_(3) weighs 5.95 g The chloride converted to the hydrous oxide and ignited to The oxide weighs 2.62 g. Calculate the percent Fe in the original mixture. Calculate the pH and pOH of the following strong base solutions:0.14 M Ba(OH)_(2) 6. Calculate the pH of the solution by adding 1 drop (0.05 ml) of 0.5 M acetic acid to 10 ml of pure water. 5.

1. To find the volume of $0.10M H_{2}SO_{4}$ needed, we can use the dilution formula: $M_1V_1 = M_2V_2$, where $M_1$ is the initial molarity, $V_1$ is the initial volume, $M_2$ is the final molarity, and $V_2$ is the final volume. In this case, $M_1 = 0.10M$, $M_2 = 0.050M$, and $V_2 = 100mL + V_1$. Solving for $V_1$, we get $V_1 = 50mL$.<br /><br />2. The balanced chemical equation for the reaction between $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ and $KMnO_{4}$ is:<br />$3KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O + 2KMnO_{4} + 3H_{2}SO_{4} \rightarrow 3K_{2}SO_{4} + 2MnSO_{4} + 6H_{2}O + 6C_{2}O_{4}$<br />From the balanced equation, we can see that 3 moles of $KHC_{2}O_{4}\cdot H_{2}C_{2}O_{4}\cdot 2H_{2}O$ react with 2 moles of $KMnO_{4}$. Therefore, the volume of $0.100M KMnO_{4}$ needed is $\frac{3}{2} \times 46.2mL = 69.3mL$.<br /><br />3. The balanced chemical equation for the reaction between ammonium chloride and sodium hydroxide is:<br />$NH_{4}Cl + NaOH \rightarrow NH_{3} + NaCl + H_{2}O$<br />From the balanced equation, we can see that 1 mole of ammonium chloride reacts with 1 mole of sodium hydroxide. The amount of sodium hydroxide remaining after the reaction is $150.0mL \times 1.00mol/L - 50.00mL \times 0.150mol/L = 0.100mol$. Therefore, the amount of ammonium chloride in the sample is also $0.100mol$. The molar mass of ammonium chloride is $53.49g/mol$, so the mass of ammonium chloride in the sample is $0.100mol \times 53.49g/mol = 5.35g$.<br /><br />4. The balanced chemical equation for the conversion of $FeCl_{3}$ to the hydrous oxide is:<br />$FeCl_{3} + 3H_{2}O \rightarrow Fe(OH)_{3} + 3HCl$<br />From the balanced equation, we can see that 1 mole of $FeCl_{3}$ produces 1 mole of $Fe(OH)_{3}$. The molar mass of $Fe(OH)_{3}$ is $106.87g/mol$, so the amount of $Fe(OH)_{3}$ produced is $\frac{2.62g}{106.87g/mol} = 0.0245mol$. Therefore, the amount of $FeCl_{3}$ in the original mixture is also $0.0245mol$. The molar mass of $FeCl_{3}$ is $162.20g/mol$, so the mass of $FeCl_{3}$ in the original mixture is $0.0245mol \times 162.20g/mol = 3.99g$. The percent Fe in the original mixture is $\frac{3.99g}{5.95g} 100\% = 67.0\%$.<br /><br />5. The pH of a solution is defined as the negative logarithm of the hydrogen ion concentration. For a strong base like $Ba(OH)_{2}$, we can use the formula $pH = 14 - pOH$, where $pOH$ is the negative logarithm of the hydroxide ion concentration. In this case, the concentration of hydroxide ions is $0.14M \times 2 = 0.28M$, so $pOH = -\log(0.28) = 0.55$. Therefore, the pH of the solution is $14 - 0.55 = 13.45$.<br /><br />6. The pH of a solution is defined as the negative logarithm of the hydrogen ion concentration. In this case, the concentration of hydrogen ions from the acetic acid is $\frac{0.05mL \