Question

Four solutions of unknown HCl concentration are titrated with solutions of NaOH. The following table lists the volume of each unknown HClsolution, the volume of NaOH solution required to reach the equivalence point, and the concentration of each NaOH solution. HCl Volume (mL) NaOH Volume (mL) [NaOH](M) 22.00 mL 31.44 mL 0.1231 M 12.00 mL 21.22 mL 0.0972 M 25.00 mL 10.88 mL 0.1088 M 3.00 mL 7.88 mL 0.1225 M

Part A Calculate the concentration (in M) of the unknown HCl solution in the first case. HCl Volume (mL) NaOH Volume (mL) [NaOH] (M) 22.00 mL 31.44 mL 0.1231 M Express your answer using four significant figures. Concentration =   M   SubmitMy AnswersGive Up Part B Calculate the concentration (in M) of the unknown HCl solution in the second case. HCl Volume (mL) NaOH Volume (mL) [NaOH] (M) 12.00 mL 21.22 mL 0.0972 M Express your answer using three significant figures. Concentration =   M   SubmitMy AnswersGive Up Part C Calculate the concentration (in M) of the unknown HCl solution in the third case. HCl Volume (mL) NaOH Volume (mL) [NaOH] (M) 25.00 mL 10.88 mL 0.1088 M Express your answer using four significant figures. Concentration =   M   SubmitMy AnswersGive Up Part D Calculate the concentration (in M) of the unknown HCl solution in the fourth case. HCl Volume (mL) NaOH Volume (mL) [NaOH] (M) 3.00 mL 7.88 mL 0.1225 M Express your answer using three significant figures. Concentration =   M

Answer 1

Acid-base neutralization:

When an acid is reacted with base creating water or solvent and an ionic salt, the reaction is known as Acid-base neutralization reaction.

Part A:

HCl Volume (mL)	NaOH Volume (mL)	[NaOH] (M)
22.00 mL	31.44 mL	0.1231 M

HCl + NaOH = NaCl + H2O

Moles of base= molarity * volume in L

= 0.1231 * 31.44/1000

= 0.00387 moles NaOH

Moles HCl = 0.00387 moles NaOH *1/1

0.00387 moles HCl

Concentration OF HCl = mole sof HCl/ volume in L

= 0.00387 moles HCl /0.022 L

= 0.176 M HCl

Part b:

HCl Volume (mL)	NaOH Volume (mL)	[NaOH] (M)
12.00 mL	21.22 mL	0.0972 M

HCl + NaOH = NaCl + H2O

Moles of base= molarity * volume in L

= 0.0972 * 21.22/1000

= 0.00206 moles NaOH

Moles HCl = 0.00206 moles NaOH *1/1

0.00206 moles HCl

Concentration OF HCl = mole sof HCl/ volume in L

= 0.00206 moles HCl /0.012 L

= 0.172 M HCl

Part C

HCl Volume (mL)	NaOH Volume (mL)	[NaOH] (M)
25.00 mL	10.88 mL	0.1088 M

HCl + NaOH = NaCl + H2O

Moles of base= molarity * volume in L

= 0.1088 * 10.88/1000

= 0.00118 moles NaOH

Moles HCl = 0.00118 moles NaOH *1/1

0.00118 moles HCl

Concentration OF HCl = mole sof HCl/ volume in L

= 0.00118 moles HCl /0.025 L

= 0.047 M HCl

Part D:

HCl Volume (mL)	NaOH Volume (mL)	[NaOH] (M)
3.00 mL	7.88 mL	0.1225 M

HCl + NaOH = NaCl + H2O

Moles of base= molarity * volume in L

= 0.1255* 7.88/1000

= 0.000989 moles NaOH

Moles HCl = 0.000989 moles NaOH *1/1

0.000989 moles HCl

Concentration OF HCl = mole sof HCl/ volume in L

= 0.000989 moles HCl /0.003 L

= 0.33 M HCl