If 36.0 mL of 0.20 M HCl is added to 30.0 mL of 0.40 M NaOH, what will be the pH of the resulting solution

The pH of the resulting solution is 12.9.

Explanation:

Molarity of the HCl solution = 0.20 M

Volume of the HCl solution = 36.0 mL

Moles of HCl = n

Molarity of NaOH solution = 0.40 M

Volume of the NaOH solution = 30.0 mL

Moles of NaOH = n'

According to the reaction, 1 mole of HCl reacts with 1 mole of NaOH then 0.0072 moles of HCl will react with :

Mole of NaOH remained unreacted = 0.012 mol - 0.0072 mol = 0.048 mol

Moles of Hydroxide ions remained unreacted = 0.0048 mol

These unreacted moles of hydroxide ion will be responsible for the pH of the final solution

Moles of hydroxide ions remained unreacted = 0.048 mol

Volume of the final solution = 36.0 mL + 30.0 mL = 66.0 mL

Molarity of the remained hydroxide ions in final solution:

The pOH of the final solution :

The pH of the resulting solution is 12.9.

pH = 1.14

Explanation:

From the above balanced reaction we observe that molar ratio for NaOH and HCl is equal.

Therefore at an equivalent point where the complete neutralization will occur, their respective concentrations are:

HCl being a strong acid:

NaOH being a strong base:

gram equivalent of HCl = 0.2 × 36 /1000 = 7.2/1000

gram equivalent of NaOH = 0.4 × 30 /1000 = 12/1000

gram equivalent of HCl remains after mixing = 4.8 /1000

Volume of mixture = 36 + 30 = 66 mL = 66/1000 litre

Normality of HCl in mixture = 4.8/1000 ÷ 66/1000 = 0.073

ion concentration = 0.073

pH = - log 0.073

pH = 1.14

i'm assuming that the correct answer is indeed the first one/ "a"..