Four solutions of unknown NaOH concentration are titrated with solutions of HCl.

Question

Four solutions of unknown NaOH concentration are titrated with solutions of HCl. The following table lists the volume of each unknown NaOH solution, the volume of HCl solution required to reach the equivalence point, and the concentration of each HCl solution. NaOH Volume (mL) HCl Volume (mL) [HCl] (M) 4.00 mL 8.17 mL 0.1699 M 15.00 mL 10.34 mL 0.1111 M 13.00 mL 10.45 mL 0.0589 M 27.00 mL 40.18 mL 0.1221 M Part A Calculate the concentration (in M) of the unknown NaOH solution in the first case.
NaOH Volume (mL) HCl Volume (mL) [HCl] (M) 4.00 mL 8.17 mL 0.1699 M Express your answer using three significant figures. Concentration =
M SubmitGive Up Part B Calculate the concentration (in M) of the unknown NaOH solution in the second case.
NaOH Volume (mL) HCl Volume (mL) [HCl] (M) 15.00 mL 10.34 mL 0.1111 M Express your answer using four significant figures. Concentration =
M SubmitGive Up Part C Calculate the concentration (in M) of the unknown NaOH solution in the third case.
NaOH Volume (mL) HCl Volume (mL) [HCl] (M) 13.00 mL 10.45 mL 0.0589 M Express your answer using three significant figures. Concentration =
M SubmitGive Up Part D Calculate the concentration (in M) of the unknown NaOH solution in the fourth case.
NaOH Volume (mL) HCl Volume (mL) [HCl] (M) 27.00 mL 40.18 mL 0.1221 M Express your answer using four significant figures. Concentration =
M SubmitGive Up
Four solutions of unknown NaOH concentration are titrated with solutions of HCl. The following table lists the volume of each unknown NaOH solution, the volume of HCl solution required to reach the equivalence point, and the concentration of each HCl solution. NaOH Volume (mL) HCl Volume (mL) [HCl] (M) 4.00 mL 8.17 mL 0.1699 M 15.00 mL 10.34 mL 0.1111 M 13.00 mL 10.45 mL 0.0589 M 27.00 mL 40.18 mL 0.1221 M Part A Calculate the concentration (in M) of the unknown NaOH solution in the first case.
NaOH Volume (mL) HCl Volume (mL) [HCl] (M) 4.00 mL 8.17 mL 0.1699 M Express your answer using three significant figures. Concentration =
M SubmitGive Up Part B Calculate the concentration (in M) of the unknown NaOH solution in the second case.
NaOH Volume (mL) HCl Volume (mL) [HCl] (M) 15.00 mL 10.34 mL 0.1111 M Express your answer using four significant figures. Concentration =
M SubmitGive Up Part C Calculate the concentration (in M) of the unknown NaOH solution in the third case.
NaOH Volume (mL) HCl Volume (mL) [HCl] (M) 13.00 mL 10.45 mL 0.0589 M Express your answer using three significant figures. Concentration =
M SubmitGive Up Part D Calculate the concentration (in M) of the unknown NaOH solution in the fourth case.
NaOH Volume (mL) HCl Volume (mL) [HCl] (M) 27.00 mL 40.18 mL 0.1221 M Express your answer using four significant figures. Concentration =
M SubmitGive Up
Four solutions of unknown NaOH concentration are titrated with solutions of HCl. The following table lists the volume of each unknown NaOH solution, the volume of HCl solution required to reach the equivalence point, and the concentration of each HCl solution. NaOH Volume (mL) HCl Volume (mL) [HCl] (M) 4.00 mL 8.17 mL 0.1699 M 15.00 mL 10.34 mL 0.1111 M 13.00 mL 10.45 mL 0.0589 M 27.00 mL 40.18 mL 0.1221 M Part A Calculate the concentration (in M) of the unknown NaOH solution in the first case.
NaOH Volume (mL) HCl Volume (mL) [HCl] (M) 4.00 mL 8.17 mL 0.1699 M Express your answer using three significant figures. Concentration =
M SubmitGive Up Part B Calculate the concentration (in M) of the unknown NaOH solution in the second case.
NaOH Volume (mL) HCl Volume (mL) [HCl] (M) 15.00 mL 10.34 mL 0.1111 M Express your answer using four significant figures. Concentration =
M SubmitGive Up Part C Calculate the concentration (in M) of the unknown NaOH solution in the third case.
NaOH Volume (mL) HCl Volume (mL) [HCl] (M) 13.00 mL 10.45 mL 0.0589 M Express your answer using three significant figures. Concentration =
M SubmitGive Up Part D Calculate the concentration (in M) of the unknown NaOH solution in the fourth case.
NaOH Volume (mL) HCl Volume (mL) [HCl] (M) 27.00 mL 40.18 mL 0.1221 M Express your answer using four significant figures. Concentration =
M SubmitGive Up

Explanation / Answer

Neutralisation reaction:

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

Number of moles of acid = number of moles of base = 1

Neutralisation formula,

M1V1/n1 = M2V2/n2

PART (A)

M1 * 4.00 = 0.1699 * 8.17

M1 = Molarity of NaOH = 0.347 M

PART (B)

M1 * 15.00 = 0.1111 * 10.34

M1 = Molarity of NaOH = 0.07658 M

PART (C)
M1 * 13.00 = 0.0589 * 10.45  

M1 = Molarity of NaOH = 0.0473 M

PART (D)

M1 * 27.00 = 0.1221 * 40.18

M1 = Molarity of NaOH = 0.1817 M

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

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