Question

titration calculation help please!!!!

a solution is made up by dissolving 5.00g of impure sodium hydroxide in water and making it up to 1.00dm3 of solution. 25cm3 of this solution is neutralized by 30.3cm3 of hydrochloric acid, of concentration 0.102mol/dm3 . calculate the percentage purity of the sodium hydroxide .

Answer 1

first of all, start with the equation
\[NaOH + HCL \rightarrow NaCL + H _{2}O\]
Find the number of moles of HCl required for titration.
\[number of moles of HCl used = \frac{ 30.3 }{ 1000 } \times 0.102 = 0.0030906\]
Use the mole ratio to find the number of moles of NaOH present in the 25 ml aliquot.
\[NaOH : HCl \]
Hence number of moles of NaOH present in the 25ml aliquot = 0.0030906
Now, this 25ml aliquot was pipetted from a 1 litre mixture.
so, you have to do a scaling since the concentration of NaOH in the 25ml and the 1litre are the same.
\[no. of moles of NaOH inthe 1 litre mixture = \frac{ 1000 }{ 25 } \times 0.0030906 = 0.123624\]
This is the number of moles of NaOH present in the 5.00g impure NaOH sample since dilution does not change the number of moles.
Use the Mr of NaOH to find the mass of NaOH in the impure sample.
22.98977 + 15.9994 + 1.00794 = 39.99711 g / mol
mass of NaOH in impure sample = 39.99711 x 0.123624 = 4.9446g
\[percentage purity =\frac{ 4.9446 }{ 5.00 } \times 100 = 98.9 percent\]

Answer 2

thank you but if you round off throughout by 3 sig figs then answer would be 100% . is that possible. can 100% be the answer