Question 4 21.50 mL of 0.320 M NaOH was added to 20.00 mL of 0.410 M HCI. The ca

Question

Question 4 21.50 mL of 0.320 M NaOH was added to 20.00 mL of 0.410 M HCI. The calculated pH (2 dec places) Thus the concentration of hydronium is r (2 sig figs) x 10^ (integer) And the concentration of hydroxide is E (2 sig figs) x 1O^ (integer) Question 5 In change from the lab manual, rather than pouring the excess NaOH and titration mixtures down the drain at the end of the period, you will pour them into a waste jar. This is due to the fact that the pH of the solutions will be greater than 11 and we shouldn't pour things down the drain that have a pH outside of the range of 4-10. Should we pour 0.00100 M HCI down the drain? (yes/no)

Explanation / Answer

Volume of NaOH, V = 21.50 mL

Concentration of NaOH, M = 0.320 M

Hence millimoles of NaOH taken = MxV = 0.320Mx21.50 mL = 6.88 millimole

Volume of HCl, V = 20.00 mL

Concentration of HCl, M = 0.410 M

Hence millimoles of HCl taken = MxV = 0.410Mx20.00 mL = 8.2 millimole

The neutralization reaction is

HCl + NaOH ---- > NaCl + H2O

1 mol, 1mol, ------- 1mol

For the above balanced reaction

1 mole of HCl reacts with 1 mole NaOH. Hence NaOH acts as limiting reactant and exhausted completely.

Hence millimoles of HCl remained = 8.2 – 6.88 = 1.32 millimole

Total volume of the solution, Vt = 21.5 mL + 20.0 mL = 41.5 mL

Hence concentration of HCl = millimoles of HCl / Vt(mL) = 1.32 millimol / 41.5 mL

= 0.0318 M

pH = - log[H+] = - log(0.0318) = (answer)

Hence concentration of hydronium ion(H3O+) = [H+] = 0.0318M

= 3.2x10-2 M (answer)

Concentration of hydroxide ion, [OH-] = 10-14 / [H+] = 10-14 / 0.0318

= 3.1x10-13 M(answer)

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.