Question

If I add acid to a 100 mL of a 0.15 M NaOH solution until it is titrated with 150 mL of acid, what will the molarity of the acid solution be?

(last question, I swear this time)

Answer 1

Assuming the acid used for titration is a strong acid (such as HCl), the balanced chemical equation for the reaction would be:

HCl + NaOH → NaCl + H2O

From the equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH. Therefore, the number of moles of HCl used for titration can be calculated as:

moles of HCl = moles of NaOH = M x V x n

where M is the molarity of NaOH, V is the volume of NaOH used (100 mL or 0.1 L), and n is the number of moles of NaOH per liter of solution (1 mole/L).

moles of HCl = 0.15 M x 0.1 L x 1 mol/L = 0.015 mol

Since the volume of acid used for titration is 150 mL or 0.15 L, we can calculate the molarity of the acid as:

Molarity of acid = moles of acid / volume of acid used

Molarity of acid = 0.015 mol / 0.15 L = 0.1 M

Therefore, the molarity of the acid solution is 0.1 M.