Titration of a Strong Acid with a Strong Base Example: HCI (strong acid) is bein

Question

Titration of a Strong Acid with a Strong Base Example: HCI (strong acid) is being titrated with NaOH (strong base). Ifyou start with 40.00 mL of 0.10 M HCI and titrate using 0.10 M NaOH, calculate the pH of the solution after the addition of (a) 0 mL, (b) 30.00 mL, (c) 40.00 mL, and (d) 45.00 mL of NaOH. Calculate the volume of NaOH required to reach the equivalence point. To do this, you must first write a balanced reaction for HCl and NaOH. How many moles of HCI are present? How many moles of NaOH wl be needed for complete neutralization? What volume of the NaOH is needed to provide this number of moles? HCI (aq) NaOH (aq) " + + 40.00 mL, 0.10 M0.10M ?# mol ?#mol Answer: 40.00 mL a Before any titrant is added (0 mL NaOH added). The solution only contains HCl, a strong acid. How do you calculate the phH of a strong acid solution? Start by writing a reaction of the strong acid in water. Answer: pH-1.00 b Before the equivalence point (30 mL NaOH added), write a balanced equation for the reaction of HCI and NaOH. How many moles of HCI were originally present? How many moles of NaOH have you added? After the HCI reacts in a 1:1 ratio with the NaOH, what has formed and what remains? HCI (aq) NaOH (aq) 40.00 mL, 0.10 M30.00 mL, 0.10 M ?#mol ?# mol

Explanation / Answer

(a) Write the balanced chemical equation for the reaction of HCl with NaOH as below.

HCl (aq) + NaOH (aq) ------> NaCl (aq) + H2O (l) …..(1)

As per the stoichiometric equation,

1 mole HCl = 1 mole NaOH.

Mole(s) of HCl in 40.00 mL of 0.10 M NaOH = (40.00 mL)*(1 L/1000 mL)*(0.10 M) = 0.004 mole.

As per the stoichiometric equation,

moles NaOH at the equivalence point = moles HCl taken = 0.004 mole.

Volume of NaOH required to reach the equivalence point = (moles of NaOH)/(molarity of NaOH) = (0.004 mole)/(0.10 M) = 0.04 L = (0.04 L)*(1000 mL/1 L) = 40.00 mL (ans).

(b) HCl is a strong acid and ionizes completely in water as below.

HCl (aq) -------> H+ (aq) + Cl- (aq)

Due to the 1:1 nature of ionization,

0.10 M HCl = 0.10 M H+.

pH = -log [H+] = -log (0.10) = 1.00 (ans).

(c) Refer to equation (1) in part (a) above.

Mole(s) NaOH added = moles HCl neutralized = (30.00 mL)*(1 L/1000 mL)*(0.10 M) = 0.003 mole.

Mole(s) HCl retained = (0.004 – 0.003) mole = 0.001 mole.

Since HCl is a strong acid, the pH of the solution is guided by HCl. The total volume of the solution is (40.00 + 30.00) mL = 70.00 mL = (70.00 mL)*(1 L/1000 mL) = 0.07 L.

Concentration of HCl = concentration of H+, i.e, [H+] = (0.001 mole)/(0.07 L) = 0.01428 mol/L = 0.01428 M.

pH = -log [H+] = -log (0.01428) = 1.845 1.85 (ans).

(d) We have already determine in (a) above that the number of moles of NaOH required to reach the equivalence point is 0.004 mole (ans).

NaCl and H2O are formed at the end of the titration. NaCl is the salt of a strong acid (HCl) and strong base (NaOH) and hence, is neutral. H2O is neutral and has pH = 7.00; therefore, the pH at the equivalence point is 7.00 (ans).

(e) We have volume of excess NaOH added (refer to part a above) = (45.00 – 40.00) mL = 5.00 mL = (5.00 mL) and hence mole(s) of excess NaOH = (5.00 mL)*(1 L/1000 mL)*(0.10 M) = 0.0005 mole.

Total volume of the solution is (40.00 + 45.00) mL = 85.00 mL = (85.00 mL)*(1 L/1000 mL) = 0.085 L.

Since NaOH is a strong base (ionizes completely to produce OH-), we must have

[OH-] = (0.0005 mole)/(0.085 L) = 5.8823*10-3 M.

We determine pOH =-log [OH-] = -log (5.8823*10-3) = 2.2304 and hence

pH = 14 – pOH = 14 – 2.2304 = 11.7696 11.77 (ans).

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