DATA TABLE MEASUREMENTS Concentration of NaOH O moV/L / 1 mol/l. Initial reading

Question

DATA TABLE MEASUREMENTS Concentration of NaOH O moV/L / 1 mol/l. Initial reading of base buret mL Final reading of base buret 29m Volume of NaOH used #2 #3 mL mL. mL. ml. mL ml Final reading of acid buret0ml ml 2ml. ml. uret Volume of HCI usedml n each titration. 2. Determine the number of moles of HCI used in each titration. 3. Determine the concentration of the HCI obtained from each titration. Then average the results. 4. If 27.3 mL. of 0.21 M NaOH is able to neutralize 37.4 mL of HCI, what is the concentration of the acid? 5. What volume of 0.12 M HCI is needed to neutralize 28.7 mL of 0.14 M KOH? Modified from: Whitman, R.L. et al. Chemistry Today 1: Laboratory Manual. 1989. Prentice-Hall Canada Inc., Scarborough

Explanation / Answer

1)

number of moles of NaOH in trial-1 = 0.1 * 21.9 * 10^-3 = 0.00219

number of moles of NaOH in trial-2 = 0.1 * 22.3 * 10^-3 = 0.00223

number of moles of NaOH in trial-3 = 0.1 * 22.5 * 10^-3 = 0.00225

2)

NaOH + HCl ------> NaCl + H2O

1 mole of HCl requires 1 mole of NaOH

number of moles of HCl in trial-1 = 0.00219

number of moles of HCl in trial-2 = 0.00223

number of moles of HCl in trial-3 = 0.00225

3)

Concentration of HCl in trial-1 = 0.00219 * 1000/20 = 0.1095 M

Concentration of HCl in trial-2 = 0.00223 * 1000/20 = 0.1115 M

Concentration of HCl in trial-3 = 0.00225 * 1000/20 = 0.1125 M

Average Concentration of HCl = (0.1095 + 0.1115 + 0.1125)/3 = 0.1112 M

4)

M1V1 = M2V2

0.21 * 27.3 = M2 * 37.4

M2 = concentration of Acid = 0.1533 M

5)

M1V1 = M2V2

0.14 * 28.7 = 0.12 * V2

V2 = volume of Acid = 33.48 mL

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