Question

If 20.60 ml of 0.0100 m aqueous hcl is required to titrate 30.00 ml of an aqueous solution of naoh to the equivalence point, what is the molarity of the naoh solution?

Answer 1

Answer : The molarity of the NaOH solution is, 0.00687 M.

Explanation :

The balanced chemical reaction will be:

$HCl+NaOH\rightarrow NaCl+H_2O$

Using neutralization method :

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=0.0100M\\V_1=20.60mL\\n_2=1\\M_2=?\\V_2=30.00mL$

Putting values in above equation, we get:

$1\times 0.0100M\times 20.60mL=1\times M_2\times 30.00mL\\\\M_2=0.00687M$

Hence, the molarity of the NaOH solution is, 0.00687 M.

Answer 2

The balanced equation for the above reaction is as follows;
NaOH + HCl -- > NaCl + H₂O
stoichiometry of NaOH to HCl is 1:1
number of HCl moles reacted - 0.0100 mol/L x 0.02060 L = 0.000206 mol 
according to 1:1 molar ratio
number of NaOH moles reacted - 0.000206 mol 
number of NaOH moles in 30.00 mL - 0.000206 mol 
therefore number of NaOH moles in 1000 mL - 0.000206 mol / 0.03000 L = 0.00687 mol
molarity of NaOH is 0.00687 M