2. The balanced equation is [latex]{\text{H}}_{2}{\text{SO}}_{4}\text{(}aq\text{)}+2\text{NaOH}\text{(}aq\text{)}\rightarrow{\text{Na}}_{2}{\text{SO}}_{4}\text{(}aq\text{)}+2{\text{H}}_{2}\text{O}\text{(}l\text{)}[/latex]

The steps to follow in solving this problem if we use volumes in milliliters are

[latex]\text{Volume NaOH}\rightarrow\text{mmol NaOH}\rightarrow{\text{mmol H}}_{2}{\text{SO}}_{4}\rightarrow{\text{M H}}_{2}{\text{SO}}_{4}[/latex]

[latex]\begin{array}{l} 1.7\cancel{\text{mL}}\times \frac{0.0811\text{mmol NaOH}}{\cancel{\text{mL}}}=0.138\text{mmol NaOH}\\ {\text{mmol H}}_{2}{\text{SO}}_{4}=0.138\cancel{\text{mmol NaOH}}\times \frac{1{\text{mmol H}}_{2}{\text{SO}}_{4}}{2\cancel{\text{mmol NaOH}}}=\text{0.069 mmol}\\ M{\text{H}}_{2}{\text{SO}}_{4}=\frac{0.069{\text{mmol H}}_{2}{\text{SO}}_{4}}{\text{20.0 mL}}=3.4\times {10}^{-3}M\end{array}[/latex]

4. In this exercise, the volume is left in units of milliliters and the number of moles is expressed in units of millimoles to compensate for the factor of 1000 difference between units. This technique is often useful in calculations. The steps involved in solving the problem are

[latex]\text{Volume Hg}{\text{(}{\text{NO}}_{3}\text{)}}_{2}\rightarrow\text{mmol Hg}\text{(}{\text{NO}}_{3}\text{)}\rightarrow{\text{mmol Cl}}^{-}\rightarrow{M\text{Cl}}^{-}[/latex]

[latex]\text{mmol Hg}{\text{(}{\text{NO}}_{3}\text{)}}_{2}=1.46\cancel{\text{mL}}\times \text{(}8.25\times {10}^{-4}\text{mmol/}\cancel{\text{mL}}\text{)}=1.20\times {10}^{-3}\text{mmol}[/latex]

[latex]{\text{mmol Cl}}^{-}=\left[1.20\times {10}^{-3}\cancel{\text{mmol Hg}{\text{(}{\text{NO}}_{3}\text{)}}_{2}}\right]\times \frac{2{\text{mmol Cl}}^{-}}{1\cancel{\text{mmol Hg}{\text{(}{\text{NO}}_{3}\text{)}}_{2}}}=2.41\times {10}^{-3}{\text{mmol Cl}}^{-}[/latex]

[latex]{M\text{Cl}}^{-}=\frac{2.41\times {10}^{-3}\text{mmol}}{0.25}=9.6\times {10}^{-3}M[/latex]

6. The reaction is [latex]{\text{GaBr}}_{3}\text{(}aq\text{)}+3{\text{AgNO}}_{3}\text{(}aq\text{)}\rightarrow 3\text{AgBr}\text{(}s\text{)}+\text{Ga}{\text{(}{\text{NO}}_{3}\text{)}}_{3}\text{(}aq\text{)}\text{.}[/latex]
Begin by considering the definition of mass percentage:

[latex]\%\text{Ga}=\frac{\text{g Ga}}{{\text{g GaBr}}_{3}}\times 100\%[/latex]

Computing this concentration will require the following approach:

[latex]\text{g AgBr}\rightarrow\text{mol AgBr}\rightarrow\text{mol Ga}{\text{(}{\text{NO}}_{3}\text{)}}_{3}\rightarrow\text{g Ga}[/latex]

Using the provided data yields

[latex]\text{mol Ga}=0.299\cancel{\text{g AgBr}}\times \frac{1\cancel{\text{mol}}}{187.7722\cancel{\text{g AgBr}}}\times \frac{1\cancel{\text{mol Ga}{\text{(}{\text{NO}}_{3}\text{)}}_{3}}}{3\cancel{\text{mol AgBr}}}\times \frac{1\cancel{\text{mol Ga}}}{1\cancel{\text{mol Ga}{\text{(}{\text{NO}}_{3}\text{)}}_{3}}}\times \frac{\text{69.723 g}}{1\cancel{\text{mol Ga}}}=3.701\times {10}^{-2}\text{g}[/latex]

Finally, the gallium mass percentage is calculated as

[latex]\%\text{Ga}=\frac{3.701\times {10}^{-2}\cancel{\text{g Ga}}}{0.165\cancel{{\text{g Ga}}_{3}}{}_{3}}=100\%=22.4\%[/latex]

8. Calculate the mass of B in the 0.063-g sample of B₂O₃. The difference of the mass of this boron and the 0.025-g sample of boron and hydrogen gives the mass of the hydrogen present. Determine the moles of B and H in the sample. Divide by the smaller number of moles to find the empirical formula. Divide the mass of the empirical formula into the assumed molecular mass of ~28 amu. That number multiplied by the subscripts of the empirical formula gives the molecular formula.
[latex]\begin{array}{l}{\text{mass of B in B}}_{2}{\text{O}}_{3}=\frac{2\times 10.811{\text{g mol}}^{-1}\text{B}}{69.6202{\text{g mol}}^{-1}{\text{B}}_{2}{\text{O}}_{3}}\times 0.063{\text{g B}}_{2}{\text{O}}_{3}=\text{0.0196 g}\end{array}[/latex]

mass of H in B₂O₃ = 0.025 g B & H – 0.0196 g B = 0.0054 g H

[latex]\begin{array}{l}\text{mol B}=\frac{\text{0.0196 g}}{10.811{\text{g mol}}^{-1}}=\text{0.00181 mol}\\ \text{mol H}=\frac{\text{0.0054 g}}{1.00794{\text{g mol}}^{-1}}=\text{0.00535 mol}\\ \text{mole ratio:}1\text{B to}\frac{0.00535\text{mol H}}{0.00181\text{mol B}}=2.96\end{array}[/latex]

Because of rounding errors, this calculation gives a ratio of 1:3. Therefore, the empirical formula is BH₃, which has a molecular mass of ~13.8 amu. Multiplication of this value by 2 gives 27.6 amu, a number of very close to the approximate mass.
Consequently, the molecular formula is B₂H₆.

10. The outline of three steps is as follows:

Convert the grams of potassium hydrogen phosphate to moles of K₂HPO₄ present; use potassium hydrogen phosphate.

Convert the balanced equation to convert moles of potassium hydrogen phosphate to moles of HNO₃; then use nitric acid.

Convert the given moles of HNO₃ to calculate the volume needed in milliliters of nitric acid.
[latex]0.2352{\text{g K}}_{2}{\text{HPO}}_{4}\times \frac{1{\text{mol K}}_{2}{\text{HPO}}_{4}}{\text{174.20 g}}\times \frac{2{\text{mol HNO}}_{3}}{1{\text{mol K}}_{2}{\text{HPO}}_{4}}\times \frac{\text{1 L}}{0.08892{\text{mol HNO}}_{3}}=\text{30.37 mL}[/latex]

All these steps can be combined into a single calculation:

[latex]\frac{0.2352{\text{g K}}_{2}{\text{HPO}}_{4}}{1}\times \frac{1{\text{mol K}}_{2}{\text{HPO}}_{4}}{\text{174.20 g}}\times \frac{2{\text{mol NHO}}_{3}}{1{\text{mol K}}_{2}{\text{HPO}}_{4}}\times \frac{1000{\text{mL HNO}}_{3}}{0.08892{\text{mol HNO}}_{3}}=30.37{\text{mL HNO}}_{3}[/latex]

12. [latex]\text{mass KOH}\rightarrow\text{mol KOH}\rightarrow{\text{mol H}}_{2}{\text{SO}}_{4}\rightarrow{\text{M H}}_{2}{\text{SO}}_{4}\rightarrow{\text{volume H}}_{2}{\text{SO}}_{4}[/latex]
[latex]\text{mol KOH}=24.74\cancel{\text{g}}\text{KOH}\times \frac{1\cancel{\text{mol KOH}}}{39.0983\cancel{\text{g}}}\times \frac{1{\text{mol H}}_{2}{\text{SO}}_{4}}{2\cancel{\text{mol KOH}}}=\text{0.031638 mol}[/latex]
[latex]{\text{volume H}}_{2}{\text{SO}}_{4}=\frac{\text{0.031638 mol}}{\text{0.3446 M}}=\text{0.09181 L}[/latex]

14. Determine the molar mass of Ca(OH)₂ and propionic acid.
Molar mass of Ca(OH)₂ = 74.093 g/mol
Molar mass of propionic acid = 88.106 g/mol

[latex]\text{mass of Ca}{\text{(}\text{OH}\text{)}}_{2}=25.0\cancel{\text{g P.A.}}\times \frac{1\cancel{\text{mol P.A.}}}{88.106\cancel{\text{g P.A.}}}\times \frac{1\cancel{\text{mol Ca}{\text{(}\text{OH}\text{)}}_{2}}}{\cancel{\text{mol P.A.}}}\times \frac{74.093}{\cancel{\text{mol Ca}{\text{(}\text{OH}\text{)}}_{2}}}=\text{21.0 g}[/latex]

16. [latex]\text{mass KHP}\rightarrow\text{mol KHP}\rightarrow\text{mol NaOH}\rightarrow\text{Concentration of NaOH}[/latex]

[latex]0.3420\cancel{\text{g KHP}}\times \frac{1\cancel{\text{mol KHP}}}{204.223\cancel{\cancel{\text{g KHP}}}}\times \frac{1\text{mol NaOH}}{1\cancel{\text{mol KHP}}}\times \frac{1}{\text{0.03573 L}}=4.687\times {10}^{-2}\text{M}[/latex]