(a) Titration curve for the titration of 5.00 mL 0.010 M NaOH(aq) with 0.005 M HCl(aq), indicating the pH of the initial and final solutions and the pH at the stoichiometric point. What volume of HCl has been added at

(b) the stoichiometric point

(c) the halfway point of the titration?

AT STOICHIOMETRIC POINT, THE VOLUME OF ACID ADDED IS 0.01 L

AT HALF-WAY POINT, THE VOLUME OF ACID IS 0.0050 L

Explanation:

In solving titration problems, you must remember this formula;

MaVa = MbVb

Since M a= 0.005 M

Mb = 0.010 M

Vb = 5 mL = 5 /1000 = 0.005 L

Va = unknown.

Solving for Va, we have:

Va = MbVb / Ma

Va = 0.010 * 0.005 / 0.005

Va = 0.01 L

So therefore, the volume of acid added at:

1. the stoichiometric point is 0.01 L

2. half-way point of titration is 0.01 /2 = 0.0050 L

For the pH:

Since HCl is a strong acid, it dissociate into {H30}+ ion.

First calculate the number of moles of hydronium ion

number of mole = concentration of hydronium ion {H30}+ * Volume

n = 0.005 * 0.01  = 0.00005 moles

A. At initial point of the titration, the volume of base added is 0 L

{H30]+ = n(H+)/ V = 0.00005 / 0.01 = 0.005 M

pH  = - log {0.005}

pH = 2.3

B. At the final point, since the volumes and concentrations of acid and base are the same, the pH is equal to 7.

n(H+) = n(OH^-)

pH = 7

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