What volume of a 3.5 M LiOH solution is needed to titrate 253 ml of a 2.75 M HF solution?

198.8mL

Explanation:

Step 1:

Data obtained from the question:

Molarity of base (Mb) = 3.5M

Volume of base (Vb) =..?

Molarity of acid (Va) = 2.75M

Volume of acid (Va) = 253mL

Step 2:

The balanced equation for the reaction.

HF + LiOH —> LiF + H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Step 3:

Determination of the volume of the base, LiOH needed for the reaction.

The volume of the base needed for the reaction can be obtained as follow:

MaVa /MbVb = nA/nB

2.75 x 253 / 3.5 x Vb = 1

Cross multiply

3.5 x Vb = 2.75 x 253

Divide both side by 3.5

Vb = 2.75 x 253 / 3.5

Vb = 198.8mL

Therefore, the volume of the base needed for the reaction is 198.8mL

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