Question

Titration of Vinegar Lab Report 1 /25 pts) Name: Part 1: Data and Calculations mass KHC.H.O- 4.0744 g Molecular Weight potassium hydrogen phthalate KHCHO. - [1 pt) #moles KHP = 10.5 pts) Volume of solution = 250 ml_ L10.5 pts] Concentration of KHP = Mop = _M (1 pt) Volume of KHP Solution to be titrated mL = [1 pt) # mols of KHP to be titrated [Moe XVw] = [1 pt) Rough Trial Trial 1 Trial 2 9.9 0.0 Initial Volume of NaOH (+0.1 mi) Final Volume of NaOH (+0.1 mL) Volume of NaoH added (+0.1 mL) 9.9 20.9 10.4 9.9 11.0 10.4 [1 pt] Average Volume of NaOH added in 3 trials (30.1 mL) "Initial volume is the initial reading of the burette and final volume is the reading after adding NOOH solution [4 pts) Use the average volume of NaOH added (VNOH) to determine the molar concentration of NaOH used in the titration. Use the equation which applies to the situation that exists at the end point: M V .M .V . More specifically since KHP is an acid and sodium hydroxide is a base, MeV MNOHVon should be used and solve for Mom

Answer 1

PART 1- Given mass of potassium hydrogen pthalate =4.0744g

• Molecular weight of potassium hydrogen pthalate (KHC8H4O4) =

39+1+8*12+4*1+4*16 (Mass of Potassium=39,Hydrogen=1,Carbon=12,Oxygen=16)

=39+1+96+4+64

=204 amu

• moles of KHP= mass/molecular weight

=4.0744 gm/204 amu

=0.019 mol/l

• Volume of solution = 250 mL=0.25 L
• Concentration of KHP =C=n/v

where C=concentration

n=no. of moles

v=volume

C=0.019/0.25

C=0.076 moles/L

• Volume of KHP to be titrated =250 ml=0.25 L
• moles of KHP to be titrated=

[MKHP VKHP]

=0.076 *0.25 (from the above calculation)

=0.019 mol/l

Avereage volume of NaOH added =

	ROUGH TRIAl	TRIAL 1	TRIAL 2
Initial volume OF NaOH	0.0	9.9	0.0
Final VOlume OF NaOH added	9.9	20.9	10.4
Volume OF NaOH added	9.9	11.0	10.4
AVerage volume of NaOH added in 3 trials	6.6	13.93	6.93

Average volume of the NaOH added=6.6+13.93+6.93/3

=9.15 mol/L

•    [MKHP VKHP] =[MNaOH VNaOH]

$\Rightarrow$ 0.019= [MNaOH] *[9.15]

$\Rightarrow$[MNaOH]=0.019/9.15

$\Rightarrow$[MNaOH] =0.002 mol/L