Please help me solving the question number 3 which is based on question number 2
ID: 501936 • Letter: P
Question
Please help me solving the question number 3 which is based on question number 2 provided below. Thank you
3. During a titration, a student used 31.92 mL ofa NaoH solution to neutralize 30.0 mLofo.2488 M oxalic acid solution. B. Oxalic acid is diprotic!) a. rite the balanced equation for the neutralization reaction. (See question 2 for the formula of oxalic acid) b. How many moles of oxalic acid were present in the sample? c. How many moles of NaOHugo were needed to neutralize the oxalic acid? d. What is the molar concentration of the NaOH ACID AND BASE TITRATION 2 111
Explanation / Answer
3.
2NaOH + C2H2O4 -----------------> Na2H2O4 + 2H2O
2 moles 1 mole
no of moles of oxalic acid = molarity * volume in L
= 0.2488*0.03 = 0.007464moles
from balanced equation
1 mole of C2H2O4 react with 2 moles of NaOH
0.007464 moles of C2H2O4 react with = 2*0.007464/1 = 0.014928 moles of NaOH
no of moles of NaOH = molarity * volume in L
0.014928 = molarity*0.03192
molarity = 0.014928/0.03192 = 0.4676M
2.
molarity = W*1000/G.M.Wt * volume in L
0.25 = W*1000/120*500
W = 120*500*0.25/1000 = 15g >>>>>answer
mass of C2H2o4 .2H2O = 15g