Consider the reaction between NaOH (aq) and H2SO4(aq). 1) Please write the balanced reaction equation. 2) What is the concentration of the sodium hydroxide solution if 25.0 mL of it required 15.8 mL of a 0.350 M sulfuric acid for complete reaction?3) What is the concentration of the sodium hydroxide solution if 25.0 mL of it required 15.8 mL of a 0.350 M sulfuric acid for complete reaction?
4) If 14.0 mL of a 1.250 M sodium hydroxide solution and 25.0 mL of a 1.10 M phosphoric acid were reacted, what is the theoretical mass, in g, of the water produced?

Explanation:

2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O .

2 ) 15.8 mL of .35 M sulphuric acid will contain .35 x .0158 moles of sulphuric acid

= .00553 moles

1 mole of H₂SO₄ reacts with 2 moles of NaOH

.00553 moles of  H₂SO₄ will react with 2  x .00553 moles of NaOH

= .01106 moles

.01106 moles were contained in 25 mL

so concentration of NaOH = .01106 / .025

= .4424 M .

4 )

3NaOH + H₃PO₄ = Na₃PO₄+ 3H₂O .

14 mL of 1.25 M NaOH will contain .014 x 1.25 moles = .0175 moles

25 mL of 1.1 M phosphoric acid = .025 x 1.1 moles = .0275 moles

.0275 moles  of phosphoric acid will require triple moles of NaOH , but it is in short supply . So NaOH is limiting reactant .

water formed = .0175 moles .

= .0175 x 18 grams of water

= .315 grams .