2. A student titrates 30.5 mL of a 3.50 g/L monoprotic acid (HA) with 15.0 mL...

Question

Question

Answer 1

Answer #1

A) Let the concentration of monoprotic acid=X

Volume of monoprotic acid*concentration of acid=volume of NaOH*Concentration of NaOH

30.5*X=15*.10

X=(15*.10)/30.5=.049M

Concentration of acid =. 049M

B)30.5ml=30.5/1000L=.0305L

.0305L*3.50g/L=.10675g

. 049mol=.10675g

1mol=.10675/.049=2.1786g

Molar mass of the acid=2.1686g

Answer 2

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