asked 212k views
5 votes
What is the HCl concentration if 52.0 mL of 0.350 M NaOH is required to titrate a 30.0 mL sample of the acid?

asked
User Sona
by
8.6k points

1 Answer

6 votes
Volume HCl = 52.0 mL in liters: 52.0 / 1000 => 0.052 L

Molarity HCl = ?

Number of moles NaOH :

V = 30,0 mL / 1000 => 0.03 L

Molarity = 0.350 M

n = M x V

n = 0.350 x 0.03 => 0.0105 moles of NaOH

Mole ratio :

HCl + NaOH = NaCl + H2O

1 mole HCl ---------- 1 mole NaOH
? moles HCl --------- 0.0105 moles NaOH

0.0105 x 1 / 1 => 0.0105 moles of HCl

Therefore:

M = n / V

M = 0.0105 / 0.052

= 0.201 M
answered
User Burak Gavas
by
8.4k points
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