Question

What volume of each of the following acids will react completely with 50.00 mL of 0.100 M NaOH?

$\begin{array}{l}\mathbf{a}\mathbf{.}\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{MHCI}\\ \mathbf{b}\mathbf{.}\mathbf{0}\mathbf{.}\mathbf{100}{\mathbf{MH}}_2{\mathbf{SO}}_3\left(\mathbf{twoacidichydrogens}\right)\\ \mathbf{c}\mathbf{.}\mathbf{0}\mathbf{.}\mathbf{200}{\mathbf{MH}}_3{\mathbf{PO}}_4\left(\mathbf{threeacidichydrogens}\right)\\ \mathbf{d}\mathbf{.}\mathbf{0}\mathbf{.}\mathbf{150}{\mathbf{MHNO}}_3\\ \mathbf{e}\mathbf{.}\mathbf{0}\mathbf{.}\mathbf{200}{\mathbf{MHC}}_2{\mathbf{H}}_3{\mathbf{O}}_2\left(\mathbf{oneacidichydrogen}\right)\\ \mathbf{f}\mathbf{.}\mathbf{0}\mathbf{.}\mathbf{300}{\mathbf{MH}}_2{\mathbf{SO}}_4\left(\mathbf{twoacidichydrogens}\right)\end{array}$

Short answer

The volume of the acids:

$\begin{array}{l}\mathrm{a}.0.100\mathrm{MHCI}=50.0\mathrm{mL}\\ \mathrm{b}.0.100{\mathrm{MH}}_2{\mathrm{SO}}_3=25.0\mathrm{mL}\\ \mathrm{c}.0.200{\mathrm{MH}}_3{\mathrm{PO}}_4\text{=8.33mL}\\ \mathrm{d}.0.150{\mathrm{MHNO}}_3=33.3\mathrm{mL}\\ \mathrm{e}.0.200{\mathrm{MHC}}_2{\mathrm{H}}_3{\mathrm{O}}_2\text{=25.0mL}\\ \mathrm{f}.0.300{\mathrm{MH}}_2{\mathrm{SO}}_4\text{=8.33mL}\end{array}$

Step-by-step solution

Definition of acids

An acid is asubstance that tastes sour in water solution, changes blue litmus paper to red, reacts with some metals to liberate hydrogen, reacts with bases to form salts, and speeds up chemical reactions.

a. Volume of the acid

$\begin{array}{l}\text{Molesof}{\mathrm{H}}_3{\mathrm{PO}}_4=\frac{0.141{\mathrm{molH}}_3{\mathrm{PO}}_4}{1.00{\mathrm{LH}}_3{\mathrm{PO}}_4}\times \frac{1.00{\mathrm{LH}}_3{\mathrm{PO}}_4}{1000{\mathrm{mLH}}_3{\mathrm{PO}}_4}\times 14.20{\mathrm{mLH}}_3{\mathrm{PO}}_4\\ \text{Molesof}{\mathrm{H}}_3{\mathrm{PO}}_4=2.00\times {10}^{-3}{\mathrm{molH}}_3{\mathrm{PO}}_4\end{array}$

Hence, the volume of the acid is 50.0mL.

b. Volume of the acid

$\begin{array}{l}\text{Molesof}{\mathrm{H}}_3{\mathrm{PO}}_4=\frac{0.141{\mathrm{molH}}_3{\mathrm{PO}}_4}{1.00{\mathrm{LH}}_3{\mathrm{PO}}_4}\times \frac{1.00{\mathrm{LH}}_3{\mathrm{PO}}_4}{1000{\mathrm{mLH}}_3{\mathrm{PO}}_4}\times 14.20{\mathrm{mLH}}_3{\mathrm{PO}}_4\\ \text{Molesof}{\mathrm{H}}_3{\mathrm{PO}}_4=2.00\times {10}^{-3}{\mathrm{molH}}_3{\mathrm{PO}}_4\end{array}$

Hence, the volume of the acid is 25.0mL.

c. Volume of the acid

$\begin{array}{l}\text{Molesof}{\mathrm{H}}_3{\mathrm{PO}}_4=\frac{0.141{\mathrm{molH}}_3{\mathrm{PO}}_4}{1.00{\mathrm{LH}}_3{\mathrm{PO}}_4}\times \frac{1.00{\mathrm{LH}}_3{\mathrm{PO}}_4}{1000{\mathrm{mLH}}_3{\mathrm{PO}}_4}\times 14.20{\mathrm{mLH}}_3{\mathrm{PO}}_4\\ \text{Molesof}{\mathrm{H}}_3{\mathrm{PO}}_4=2.00\times {10}^{-3}{\mathrm{molH}}_3{\mathrm{PO}}_4\end{array}$

Hence, the volume of the acid is 8.33 mL.

d. Volume of the acid

$\begin{array}{l}\text{Molesof}{\mathrm{H}}_3{\mathrm{PO}}_4=\frac{0.141{\mathrm{molH}}_3{\mathrm{PO}}_4}{1.00{\mathrm{LH}}_3{\mathrm{PO}}_4}\times \frac{1.00{\mathrm{LH}}_3{\mathrm{PO}}_4}{1000{\mathrm{mLH}}_3{\mathrm{PO}}_4}\times 14.20{\mathrm{mLH}}_3{\mathrm{PO}}_4\\ \text{Molesof}{\mathrm{H}}_3{\mathrm{PO}}_4=2.00\times {10}^{-3}{\mathrm{molH}}_3{\mathrm{PO}}_4\end{array}$

Hence, the volume of the acid is 33.3 mL.

e. Volume of the acid

$\begin{array}{l}\text{Molesof}{\mathrm{H}}_3{\mathrm{PO}}_4=\frac{0.141{\mathrm{molH}}_3{\mathrm{PO}}_4}{1.00{\mathrm{LH}}_3{\mathrm{PO}}_4}\times \frac{1.00{\mathrm{LH}}_3{\mathrm{PO}}_4}{1000{\mathrm{mLH}}_3{\mathrm{PO}}_4}\times 14.20{\mathrm{mLH}}_3{\mathrm{PO}}_4\\ \text{Molesof}{\mathrm{H}}_3{\mathrm{PO}}_4=2.00\times {10}^{-3}{\mathrm{molH}}_3{\mathrm{PO}}_4\end{array}$

Hence, the volume of the acid is 25.0mL

f. Volume of the acid

$\begin{array}{l}\text{Molesof}{\mathrm{H}}_3{\mathrm{PO}}_4=\frac{0.141{\mathrm{molH}}_3{\mathrm{PO}}_4}{1.00{\mathrm{LH}}_3{\mathrm{PO}}_4}\times \frac{1.00{\mathrm{LH}}_3{\mathrm{PO}}_4}{1000{\mathrm{mLH}}_3{\mathrm{PO}}_4}\times 14.20{\mathrm{mLH}}_3{\mathrm{PO}}_4\\ \text{Molesof}{\mathrm{H}}_3{\mathrm{PO}}_4=2.00\times {10}^{-3}{\mathrm{molH}}_3{\mathrm{PO}}_4\end{array}$

Hence, the volume of the acid is 8.33 mL.