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

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

The total ionic equation of the balanced molecular equation is like

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

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

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

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-)

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

As you know, H₂SO₄ decomposes as

${\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)鈫�2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2鈭�}\left(\mathrm{aq}\right)$

Now, you can see

1 mol H₂SO₄ produces 1 mol SO₄^2-.

Therefore, moles of H₂SO₄

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

According to the balanced molecular equation

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

2 moles of NaOH reacts with 1 mol of H₂SO₄

Therefore, moles of NaOH

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

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

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.

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.

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