Answer:
#a. OH⁻(aq) + H⁺(aq) → H₂O(l)
#b. 0.000988 moles
#c. 0.000988 moles
#d.30.875 L
Solution and explanation:
38.0 mL of a 0.026 M of HCl is needed to react completely with 0.032 M NaOH solution.
#a. Net ionic equation for the reaction
The reaction between NaOH and HCl is known as neutralization reaction which results to the formation of salt and water as the product.
There equation for the reaction is;
NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)
The net ionic equation for a reaction involves only the ions that have changed their state.
The complete ionic equation is;
Na⁺(aq)OH⁻(aq) + H⁺(aq)Cl⁻(aq) → Na⁺(aq)Cl(aq) + H₂O(l)
Therefore, the net ionic equation will be;
OH⁻(aq) + H⁺(aq) → H₂O(l)
#b. Moles of HCl originally present.
Number of moles is given by multiplying the volume of the solution by its molarity or concentration.
Moles = Volume × Concentration
Volume of HCl = 38.0 mL
Concentration = 0.026 M
Therefore;
Moles = 0.038 × 0.026
= 0.000988 moles
Moles of HCl is 0.000988 moles
#c. Moles of NaOH used in the reaction
From the equation , 1 mole of HCl reacts with 1 mole of NaOH
Therefore;
The mole ratio of HCl : NaOH is 1 : 1
Hence, moles of NaOH will be 0.000988 moles
0.000988 moles were neutralized by the acid.
#d. Volume of NaOH required to react with the acid
Moles = Volume × Molarity
Therefore;
Volume = Moles ÷ Molarity
Moles of NaOH is 0.000988 moles
Molarity of NaOH is 0.032 M
Thus;
Volume of NaOH = 0.000988 moles ÷ 0.032 M
= 0.030875 mL
= 30.875 L
Therefore, the volume of the NaOH solution required to react with the acid is 30.875 L.