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tester [92]
2 years ago
6

.950 L of .420 M H2SO4 is mixed with .900 L of .260 M KOH. What concentration of sulfuric acid remains after neutralization?

Chemistry
1 answer:
kipiarov [429]2 years ago
3 0
H₂SO₄:

V=0,95L
Cm=0,420mol/L

n = CmV = 0,42mol/L * 0,95L = 0,399mol

KOH:

V=0,9L
Cm=0,26mol/L

n = CmV = 0,26mol/L * 0,9L = 0,234mol

H₂SO₄            +           2KOH ⇒ K₂SO₄ + 2H₂O
1mol                :           2mol
0,399mol         :           0,234mol
                                    limiting reagent
reamins: 0,399mol - 0,117mol = 0,282mol

n = 0,282mol
V = 0,950L + 0,900L = 1,85L

Cm = n / V = 0,282mol / 1,85L ≈ 0,152M
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