Question

4. What is the molar concentration and grams/Liter of a NaOH solution if 86 ml are titrated to an
endpoint by 375 ml of a solution of HCl that is .0175 M?
4. g/L-
4. Molarity,

Answer 1

Answer:

0.0763 M  NaOH solution

3.05 g/mol  NaOH solution

Explanation:

NaOH + HCl ----> NaCl + H2O

1 mol      1 mol

M(NaOH)*V(NaOH) = M(HCl)*V(HCl)

M(NaOH) *86 mL = 0.0175M*375 mL

M(NaOH) = 0.0175M*375 mL/86 mL = 0.0763 M

0.0763 M = 0.0763 mol/L

M(NaOH) = 23 + 16 + 1 = 40 g/mol

0.0763 mol/L * 40 g/mol = 3.05 g/mol NaOH