What is the concentration of the NaOH solution, if in a titration, 20 mL of NaOH reacts with 19.82 mL of 0.1018M HCl ?
During the titration the reaction occurring is NaOH + HCl --> NaCl + H2O. The number of moles of NaOH in the NaOH solution is equal to the number of moles of HCl in the HCl solution.
Let the concentration of the NaOH solution be X. 20 mL of this solution reacts with 19.82 mL of 0.1018 M HCl. The number of moles of HCl is 19.82*0.1018/1000 = 0.0020 moles.
As the number of moles of NaOH is equal to this,
X*20/1000 = 0.0020
=> X = 2/20 = .1 M
The molarity of NaOH is 0.1 M approximately.
This is an acid base reaction:
HCl + NaOH -> H2O + NaCl
The mole ratio is 1:1 (1 mole of HCl to 1 mol of NaOH)
Now, we'll have to find the number of moles of reactant HCl:
n = 0.1018*0.01982
n = 0.002 mol of HCl
Now, we'll determine the concentration of NaOH, using the equality:
M(HCl)*V(HCl) = M(NaOH)*V(NaOH)
M represents the molarity of acid or base
V represents the volume of acid or base
0.002 = 0.02*M(NaOH)
M(NaOH) = 0.002/0.02
M(NaOH) = 0.1 [mol/L]