35.0 mL of acid with an unknown concentration is titrated with 24.6 mL of 0.432 M base. What is the concentration of the acid? A
ssume the acid contributes 1 mole of (H+) ions/mole of acid and the base contributes 1 mole of (OH-) ions/mole of base.
3.29 M
0.615 M
0.304 M
1 answer:
Given:
35.0 mL of acid with an unknown concentration
24.6 mL of 0.432 M base
Required:
Concentration of the acid
Solution:
M1V1 = M2V2
M1 (35.0 mL of acid)
= (0.432 M base) (24.6 mL
of base)
V1 = (0.432
M base) (24.6 mL of base) /
(35.0 mL of acid)
M1 = 0.304 M of acid
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<em>V = 151 mL = 151 cm³</em>
<em>d = 0,789 g/mL = 0,789 g/cm³</em>
--------------------------------------
d = m/V
m = d×V
m = 0,789×151
<u>m = 119,139g</u>
