Question

The flask represents the products of the titration of 25mL of sulfuric acid with 25mL of sodium hydroxide.

(a) Write balanced molecular, total ionic, and net ionic equations for the reaction.

(b) If each orange sphere represents 0.010 mol of sulfate ion, how many moles of acid and of base reacted?

(c) What are the molarities of the acid and the base?

Short answer

Answer of subpart (a):

Answer: You need to write the balanced molecular equation, total ionic equation and net ionic equation for the reaction.

Answer of subpart (b):

Answer: You need to calculate how many moles of acid and of base reacted.

Answer of subpart (c):

Answer: You need to calculate the molarity of acid and base

Step-by-step solution

Molecular equation for precipitation reaction

Titration of sulfuric acid with sodium hydroxide produces sodium sulfate and water. The molecular balanced equation is like

$2\mathrm{NaOH}\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to {\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Total ionic equation for precipitation reaction

The total ionic equation of the balanced molecular equation is like

$2{\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{OH}}_{}^{-}\left(\mathrm{aq}\right)+2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)\to 2{\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+{{\mathrm{SO}}_4^2}^{-}\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Net ionic equation for precipitation reaction

The Na⁺ and SO₄^2- ions are spectator ions. Hence, they will not be present in the net ionic equation. Therefore, the net ionic equation is like

$\begin{array}{c}2{\mathrm{OH}}_{}^{-}\left(\mathrm{aq}\right)+2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\\ {\mathrm{OH}}_{}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\end{array}$

Calculation of total moles of orange spheres

Answer of subpart (b):

Answer: You need to calculate how many moles of acid and of base reacted.

According to the question, orange sphere represents SO₄^2- ions.

Moles of each orange sphere = 0.010mol

Total no of orange sphere = 2

Hence, total moles of orange sphere (SO₄^2-)

$\begin{array}{l}=(2\times 0.010)\left(\mathrm{mol}\right)\\ =0.02\left(\mathrm{mol}\right)\end{array}$

Calculation of moles of H2SO4

As you know, H₂SO₄ decomposes as

${\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to 2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)$

Now, you can see

1 mol H₂SO₄ produces 1 mol SO₄^2-.

Therefore, moles of H₂SO₄

_{$\begin{array}{l}=\left(0.02\times \frac{1}{1}\right)\left(\mathrm{mol}\right)\\ =0.02\left(\mathrm{mol}\right)\end{array}$}

Calculation of moles of NaOH

According to the balanced molecular equation

$2\mathrm{NaOH}\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to {\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

2 moles of NaOH reacts with 1 mol of H₂SO₄

Therefore, moles of NaOH

$\begin{array}{l}\text{=}\left(\text{0.02×}\frac{\text{2}}{\text{1}}\right)\left(\text{mol}\right)\\ \text{=0.04}\left(\text{mol}\right)\end{array}$

Hence, 0.02 mole of acid reacts with 0.04 mole of base.

Calculation of molarity of H2SO4

Answer of subpart (c):

Answer: You need to calculate the molarity of acid and base.

According to the question,

Volume of H₂SO₄ = 25mL = 0.025L

Again you know,

_{$\mathrm{molarity}=\frac{\mathrm{moles}}{\mathrm{volume}}$}

Now, molarity of H₂SO₄

$\begin{array}{l}=\frac{0.02}{0.025}\left(\mathrm{M}\right)\\ =0.8\left(\mathrm{M}\right)\end{array}$

Hence, molarity of H₂SO₄ is 0.8M.

Calculation of molarity of NaOH

According to the question,

Volume of NaOH = 25mL = 0.025L

Again you know,

_{$\mathrm{molarity}=\frac{\mathrm{moles}}{\mathrm{volume}}$}

Now, molarity of NaOH

$\begin{array}{l}=\frac{0.04}{0.025}\left(\mathrm{M}\right)\\ =1.6\left(\mathrm{M}\right)\end{array}$

Hence, molarity of NaOH is 1.6M.

Conclusion

The molarity of acid is 0.8M and molarity of base is 1.6M.