III. Molecular Weight of Unknown Acid # of acidic hydrogens 3 Trial 1 Trial 2 Tr
ID: 548546 • Letter: I
Question
III. Molecular Weight of Unknown Acid # of acidic hydrogens 3 Trial 1 Trial 2 Trial 3 Weight of acid unknown Base buret reading, final Base buret reading, initial Volume of base used, ml Moles of base used 4-1 42 Knowing how many acid hydrogens there are in your unknown, write a balanced molecular èquation for the neutralization. Using this balanced equation, calculate the number of moles of your unknown acid used in each trial anc enter in the table below. Since the molecular weight of the acid weight of acid you may calculate the molecular weight of the acid determined in each trial. moles of acid Trial 1 Trial 2 Trial 3 Moles of acid titrated Molecular weight of acid Average Molecular Weight Show calculations:Explanation / Answer
SOLUTION:-
In the given problem it would be known which acid are we using and which base is being used with molarity of standard solution
So taking assumption : Acid=HCl, Base=NaOH,with 0.15 molar standard solution of sodium hydroxide (NaOH).
FOR 1st TRIAL Volume of base used = 30ml
Write down the equation of the chemical reaction that the titration is based on., it is HCl + NaOH = NaCl + H2O.
Multiply the volume (in L) of the standard solution by its concentration (in moles/L) to determine the number of moles of the titrant used for titration. the volume of the NaOH solution used to reach the endpoint was 30ml or 0.030 L For 1st Trial. Hence, the number of moles of NaOH equals 0.030 L x 0.15 moles/L, or 0.0045 moles.
To determine the number of moles of the unknown compound using the equation of its chemical reaction, it's very simple to calculate, because the equation shows that 1 mole of HCl reacts with 1 mole of NaOH. Thus, if you used 0.0045 moles of NaOH to neutralize the HCl, No. of mole for HCL is 0.0045
Now we can calculate molecular wt. of acid.
Molecular weight of Acid =Weight of acid/Moles of acid
=1g/0.0045
= 222.22g
SIMILARLY FOR 2nd TRIAL:
Volume of base used = 37 ml
Moles of HCL = 0.037 * 0.15 =0.00555
Molecular weight of Acid =Weight of acid/Moles of acid
1.01g/0.00555
=181.94 g