If 15.98 mL of 0.1080 M KOH solution reacts with 52.00 mL of HC2H302,
what is the molarity of the acid solution? (Don't forget significant figures)
KOH + HC2H302 - KC2H302 + H2O

0.0332 M

Explanation:

Step 1:

The balanced equation for the reaction.

KOH + HC2H302 -> KC2H302 + H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question.

Volume of base (Vb) = 15.98 mL

Molarity of base (Mb) = 0.1080 M

Volume of acid (Va) = 52.00 mL

Molarity of acid (Ma) =?

Step 3:

Determination of the molarity of the acid.

Using the formula:

MaVa/MbVb = nA/nB

The molarity of the acid can be obtained as follow

MaVa/MbVb = nA/nB

Ma x 52 / 0.1080 x 15.98 = 1

Cross multiply to express in linear form

Ma x 52 = 0.1080 x 15.98

Divide both side by 52

Ma = (0.1080 x 15.98) /52

Ma = 0.0332 M

Therefore, the molarity of the acid solution is 0.0332 M

the answer is barium and copper bromide