Answer :
For neutralization in a titration of 35.00 mL of 0.737 M H₂SO₄, 62.4 mL of a 0.827 M KOH solution is needed.
H₂SO₄ + 2 KOH ---> K₂SO₄+ 2 H₂O
from the concentration equation
m1 n1 v1 = m2 n2 v2 ( since m = molarity , n = factor and v = volume)
M1 = 0.737 M
n1 = 2
V1 = 35 mL
M2 = 0.827 M
n2 = 1
V2 = ??
0.737 m x 2 x 35 mL = 0.827 m x 1 x v2
v2 =62.4
Hence, 62.4 mL of a 0.827 M KOH solution is needed for neutralization.
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