A student titrated 20.0 mL of 0.410 M HCl with 0.320 M NaOH 1. calculate the vol
ID: 897680 • Letter: A
Question
A student titrated 20.0 mL of 0.410 M HCl with 0.320 M NaOH
1. calculate the volume of titrant needed to reach the equivalence point
2. Calculate the solution pH when 21.50 mL of titrant has been added.
3.What you will do with the NaOH solution remaining in your burette after you have completed your titrations?
4.When reading the procedure, you may notice that two out of the four titrations use phenolphthalein as indicator, and the other two titration uses methyl orange as indicator. What titrations require methyl orange as indicator and why was phenolphthalein replaced with methyl orange?
Explanation / Answer
1.
NaOH + HCL --> NaCL + H20
no of mol of HCL = 20*0.41=8.2 millimole
1 mol of HCL react with 1 mol of NaOH
8.2 x 10^-3 mol of HCL react with 8.2 x 10^-3 mol of NaOH
volume = 8.2 x 10^-3/0.32 = 25.625 ml.
2. no of mol of HCL = 20*0.41=8.2 millimole
no of mol of NAOH = 0.32*21.5=6.88 millimole
moles neutralized=6.88 millimoles
moles of HCL remaining=8.2-6.88=1.32 millimole
volume=20+21.5=41.5 ml
molarity of hcl=1.32/41.5=3.18 *10^-2
pH=-log(3.18 *10^-2)=1.4975
3.
drain the remaining NaOH from your buret into your 100 mL beaker. Discard all solutions in the sink with plenty of water. don't waste NaOH collect it for further titration.
4.
procedure is not given so i am giving general way to choose indicator.
In the methyl orange case, the half-way stage where the mixture of red and yellow produces an orange color happens at pH 3.7 - nowhere near neutral
The half-way stage happens at pH 9.3. so in the titration which gives acidic mixture use methyl orange and the one which is basic use phenolphthalein.
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