Question

Four solutions of unknown NaOH concentration are titrated with solutions of HCl. The following table lists the volume of each unknown NaOH solution, the volume of HCl solution required to reach the equivalence point, and the concentration of each HCl solution.

NaOH Volume (mL)	HCl Volume (mL)	[HCl] (M)
7.00 mL	9.17 mL	0.2099 M
18.00 mL	10.84 mL	0.1111 M
15.00 mL	10.75 mL	0.0489 M
26.00 mL	35.18 mL	0.1421 M

Part A: Calculate the concentration (in M) of the unknown NaOH solution in the first case.

NaOH Volume (mL)	HCl Volume (mL)	[HCl] (M)
7.00 mL	9.17 mL	0.2099 M

Part B: Calculate the concentration (in M) of the unknown NaOH solution in the second case.

NaOH Volume (mL)	HCl Volume (mL)	[HCl] (M)
18.00 mL	10.84 mL	0.1111 M

Part C: Calculate the concentration (in M) of the unknown NaOH solution in the third case.

NaOH Volume (mL)	HCl Volume (mL)	[HCl] (M)
15.00 mL	10.75 mL	0.0489 M

Part D: Calculate the concentration (in M) of the unknown NaOH solution in the fourth case.

NaOH Volume (mL)	HCl Volume (mL)	[HCl] (M)
26.00 mL	35.18 mL	0.1421 M

Answer 1

Solution:-

HCl and NaOH reacts in 1:1 ratio as is clear from the balanced equation....

HCl + NaOH ------> NaCl + H₂O

For this type of acid base titrations when they react in 1:1 mol ratio we use the simple molarity equation...

M₁V₁ = M₂V₂

Where M₁ is the molarity of HCl solution and V₁ is its volume. M₂ is the molarity of NaOH and V₂ is its volume.

Let's plug in the values....

For part (A)

0.2099M x 9.17ml = M₂ x 7.00ml

divide both sides by 7.00ml

M₂ = 0.2750M

All the other parts could also be solved in the same way..

Part (B)

0.1111M x 10.84ml = M₂ x 18.00ml

divide both sides by 18.00ml

M₂ = 0.0669M

Part (C)

0.0489M x 10.75ml = M₂ x 15.00ml

divide both sides by 15.00ml

M₂ = 0.0350M

part (D)

0.1421M x 35.18ml = M₂ x 26.00ml

divide both sides by 26.00ml

M₂ = 0.01923M