Question

1. a 110 ml solution of 0.340 m hcl(aq) is mixed with a solution of 330 ml of 0.150 m hno₃(aq). the solution is then diluted to a final volume of 1.00 l. how many moles of h⁺ are present in the final solution?
2. when 38.0 ml of 0.1250 m h₂so₄ is added to 100 ml of a solution of pbi₂, a precipitate of pbso₄ forms. the pbso₄ is then filtered from the solution, dried, and weighed. if the recovered pbso₄ is found to have a mass of 0.0471 g, what was the concentration of iodide ions in the original solution?
3. 34.62 ml of 0.1510 m naoh was needed to neutralize 50.0 ml of an h₂so₄ solution. what is the concentration of the original sulfuric acid solution?
4. a sample of 0.670 g of an unknown compound containing barium ions (ba²⁺) is dissolved in water and treated with an excess of na₂so₄. if the mass of the baso₄ precipitate formed is 0.4105 g, what is the percent by mass of ba in the original compound?
5. balance the following by the method of ½ reactions:

a. bro₃⁻ + n₂h₄ → br⁻ + n₂ + h₂o
b. cn⁻ + mno₄⁻ + oh⁻ → mno₂(s) + cno⁻ + h₂o

Explanation:

1.

Step1: Calculate moles of $H^+$ from $HCl$

Use the formula $n = M\times V$, where $n$ is moles, $M$ is molarity and $V$ is volume in liters. For $HCl$, $V_{HCl}=110\ mL = 0.110\ L$ and $M_{HCl}=0.340\ M$. So $n_{HCl}=M_{HCl}\times V_{HCl}=0.340\ mol/L\times0.110\ L = 0.0374\ mol$. Since $HCl$ is a strong - acid and dissociates completely as $HCl
ightarrow H^++Cl^-$, the moles of $H^+$ from $HCl$ is $0.0374\ mol$.

Step2: Calculate moles of $H^+$ from $HNO_3$

For $HNO_3$, $V_{HNO_3}=330\ mL = 0.330\ L$ and $M_{HNO_3}=0.150\ M$. Using $n = M\times V$, we get $n_{HNO_3}=M_{HNO_3}\times V_{HNO_3}=0.150\ mol/L\times0.330\ L = 0.0495\ mol$. Since $HNO_3$ is a strong - acid and dissociates completely as $HNO_3
ightarrow H^++NO_3^-$, the moles of $H^+$ from $HNO_3$ is $0.0495\ mol$.

Step3: Total moles of $H^+$ in the final solution

The total moles of $H^+$ in the final solution is the sum of moles of $H^+$ from $HCl$ and $HNO_3$. So $n_{H^+}=n_{HCl}+n_{HNO_3}=0.0374\ mol + 0.0495\ mol=0.0869\ mol$.

Step1: Calculate moles of $H_2SO_4$ added

Use the formula $n = M\times V$. Given $V = 38.0\ mL=0.0380\ L$ and $M = 0.1250\ M$. So $n_{H_2SO_4}=0.1250\ mol/L\times0.0380\ L = 0.00475\ mol$.

Step2: Calculate moles of $PbSO_4$ formed

The molar mass of $PbSO_4$ is $M_{PbSO_4}=207.2 + 32.07+4\times16.00=303.27\ g/mol$. Given $m_{PbSO_4}=0.0471\ g$. Using $n=\frac{m}{M}$, we get $n_{PbSO_4}=\frac{0.0471\ g}{303.27\ g/mol}=0.000155\ mol$.

Step3: Determine moles of $PbI_2$ reacted

From the reaction $PbI_2 + H_2SO_4
ightarrow PbSO_4+2HI$, the mole - ratio of $PbI_2$ to $PbSO_4$ is $1:1$. So $n_{PbI_2}=n_{PbSO_4}=0.000155\ mol$.

Step4: Calculate moles of $I^-$ in the original solution

Since $1\ mol$ of $PbI_2$ contains $2\ mol$ of $I^-$, $n_{I^-}=2\times n_{PbI_2}=2\times0.000155\ mol = 0.00031\ mol$.

Step5: Calculate the concentration of $I^-$ in the original solution

The volume of the $PbI_2$ solution is $V = 100\ mL = 0.100\ L$. Using $M=\frac{n}{V}$, we get $M_{I^-}=\frac{0.00031\ mol}{0.100\ L}=0.0031\ M$.

Step1: Write the balanced chemical equation

The reaction between $NaOH$ and $H_2SO_4$ is $2NaOH + H_2SO_4
ightarrow Na_2SO_4+2H_2O$. The mole - ratio of $NaOH$ to $H_2SO_4$ is $2:1$.

Step2: Calculate moles of $NaOH$ used

Use the formula $n = M\times V$. Given $V_{NaOH}=34.62\ mL = 0.03462\ L$ and $M_{NaOH}=0.1510\ M$. So $n_{NaOH}=0.1510\ mol/L\times0.03462\ L = 0.005228\ mol$.

Step3: Calculate moles of $H_2SO_4$ reacted

From the mole - ratio, $n_{H_2SO_4}=\frac{1}{2}n_{NaOH}=\frac{1}{2}\times0.005228\ mol = 0.002614\ mol$.

Step4: Calculate the concentration of $H_2SO_4$

The volume of $H_2SO_4$ is $V_{H_2SO_4}=50.0\ mL = 0.0500\ L$. Using $M=\frac{n}{V}$, we get $M_{H_2SO_4}=\frac{0.002614\ mol}{0.0500\ L}=0.05228\ M$.

Answer:

$0.0869\ mol$

2.