Answer to Question #100482 in General Chemistry for Malathy

Question

B.ANALYSIS OF AN UNKNOWN ACID
TRIAL 1
a)Mass of bottle +unknown = 108.41g
b)Mass of bottle = 108.1917g
c)Mass of unknown used = 0.22g
d)Final buret reading = 13.1ml
e)Initial buret reading =0ml
f)mL of NaOH used =13.1ml
g)Mass of KHP in unknown=______?
h)Percent of KHP in unknown=______?

TRIAL2
a)Mass of bottle +unknown = 97.616g
b)Mass of bottle = 97.42g
c)Mass of unknown used = 0.196g
d)Final buret reading = 23.2ml
e)Initial buret reading =13.1
f)mL of NaOH used =13.1ml
g)Mass of KHP in unknown=______?
h)Percent of KHP in unknown=______?

TRIAL3
a)Mass of bottle +unknown = 111.9953g
b)Mass of bottle = 111.7953g
c)Mass of unknown used = 0.20g
d)Final buret reading = 33.0ml
e)Initial buret reading =23.2ml
f)mL of NaOH used =9.8ml
g)Mass of KHP in unknown=______?
h)Percent of KHP in unknown=______?
●Calculate Average Percent of KHP =_________?
●Standard deviation=_________?

Answer 1

Trial 1

moles of NaOH used = moles of KHP

moles of NaOH used = (Volume of NaOH used)*(Concentration of NaOH)

You have not specified a concentration, so be it, concentration of NaOH = 0.05 mol/L. If the concentration is different, it must be replaced.

moles of NaOH used = (0.0131 L)*(0.05 mol/L) = 0.000655 moles

moles of NaOH used = moles of KHP = 0.000655 moles

mass of KHP in unknown = moles of KHP*(FW of KHP) = (0.000655 moles)*(204.2212 g/mol) = 0.1338 g KHP

% KHP in unknown = (mass KHP in unknown)*100%/weight of unknown

= (0.1338g/0.22g)*100% = 60.82 %

Trial 2

moles of NaOH used = (0.0131 L)*(0.05 mol/L) = 0.000655 moles

moles of NaOH used = moles of KHP = 0.000655 moles

mass of KHP in unknown = moles of KHP*(FW of KHP) = (0.000655 moles)*(204.2212 g/mol) = 0.1338 g KHP

% KHP in unknown = (mass KHP in unknown)*100%/weight of unknown

= (0.1338g/0.196g)*100% = 68.27 %

Trial 3

moles of NaOH used = (0.0098 L)*(0.05 mol/L) = 0.00049 moles

moles of NaOH used = moles of KHP = 0.00049 moles

mass of KHP in unknown = moles of KHP*(FW of KHP) = (0.00049 moles)*(204.2212 g/mol) = 0.1001 g KHP

% KHP in unknown = (mass KHP in unknown)*100%/weight of unknown

= (0.1001g/0.20g)*100% = 50.05 %

Average Percent of KHP = 59.71 %

Standard deviation = 9.46