Question

find average molarity of the NaOH solution and avg deviation

Answer 1

You have used oxalic acid, which is a diprotic acid, to titrate NaOH, which is a monobase. The reaction between them is:

$C_{2}O_{4}H_{2}+2NaOH\rightarrow C_{2}O_{4}^{2-}+2Na^{+}+2H_{2}O$

Which means that per each mole of acid that has reacted, TWO moles of base reacted. This means that in the titration, the number of moles of NaOH in each trial is given by double the number of moles of oxalic acid; that is:

Titration 1, 2, 3 and 4: 0.0258 moles of NaOH

To calculate the molar concentration corresponding to each case, we must divide by the volume in liters, since the definition of molarity is moles per liter:

$1:M=\frac{n}{V(L)}=\frac{0.0258moles}{0.0108L}=2.39M$

$2:M=\frac{n}{V(L)}=\frac{0.0258moles}{0.0092L}=2.80M$

$3:M=\frac{n}{V(L)}=\frac{0.0258moles}{0.0074L}=3.49M$

$4:M=\frac{n}{V(L)}=\frac{0.0258moles}{0.0086L}=3.00M$

The average is given by the sum of all the values divided by the number of values (4):

$\overline{M}=\frac{2.39M+2.80M+3.49M+3.00M}{4}=2.92M$

And the standard deviation is given by the sum of the squares of the differences between each value and the mean value, divided by N-1 and then all inside a square root. In this case, it would be:

$\sigma _{M}=\sqrt{\frac{1}{N-1}\cdot [(2.39M-2.92M)^{2}+(2.80M-2.92M)^{2}+(3.49M-2.92M)^{2}+(3.00M-2.92M)^{2}]}=0.46M$