Question #12648

0.5 L of a 0.15 M HCI solution is titrated with a solution of 0.30M Naoh?

A) what is the Ph before addition of Naoh?

B) what is the total Number of moles of acid?

C) what is the Ph after addition of 125 ML of Naoh solution ?

D) what Volume of the Naoh solution is required to reach the equivalence point ?

E) what is the Ph at the equivalence point ?

F) What is the Ph after addition of 375ml (total) of Naoh solution?

A) what is the Ph before addition of Naoh?

B) what is the total Number of moles of acid?

C) what is the Ph after addition of 125 ML of Naoh solution ?

D) what Volume of the Naoh solution is required to reach the equivalence point ?

E) what is the Ph at the equivalence point ?

F) What is the Ph after addition of 375ml (total) of Naoh solution?

Expert's answer

a. pH = -log10[H+] = -log(0.15) = 0.824

b. Moles of acid = 0.5 L x 0.15 mol/

L = 0.075 mol

c. 125 mL NaOH solution added, i.e. 0.125 L x 0.30 mol L-1 =

0.0375 mol NaOH; this reacts quantitatively with the HCl, which is in excess

(A), but now there are 0.5L + 0.125 L of volume; so [H+] will diminish and pH

increase.

d. We need to add 0.075 mol NaOH to neutralize the starting

solution 0.075 mol/0.3 mol L-1=0.25 L

e. At equivalence we have a solution

that is stoichiometric in NaCl, the salt of a strong base and strong acid. The

pH 7

f.NaOH added is 0.375 L x 0.3 mol L-1 = 0.1125 mol NaOH; 0.075 mol of

this reacts with the HCl; so 0.1125 - 0.075=0.0375 mol NaOH in excess. pOH (

-log10[OH-]) and pH = 14 - pOH=14-1.43=12.57

b. Moles of acid = 0.5 L x 0.15 mol/

L = 0.075 mol

c. 125 mL NaOH solution added, i.e. 0.125 L x 0.30 mol L-1 =

0.0375 mol NaOH; this reacts quantitatively with the HCl, which is in excess

(A), but now there are 0.5L + 0.125 L of volume; so [H+] will diminish and pH

increase.

d. We need to add 0.075 mol NaOH to neutralize the starting

solution 0.075 mol/0.3 mol L-1=0.25 L

e. At equivalence we have a solution

that is stoichiometric in NaCl, the salt of a strong base and strong acid. The

pH 7

f.NaOH added is 0.375 L x 0.3 mol L-1 = 0.1125 mol NaOH; 0.075 mol of

this reacts with the HCl; so 0.1125 - 0.075=0.0375 mol NaOH in excess. pOH (

-log10[OH-]) and pH = 14 - pOH=14-1.43=12.57

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