Titration of Analytical Sample

An analytical sample of NaOH sample was titrated with 31.5 mL of 0.80 M HCl in order to reach the equivalence point. Determine the concentration of NaOH sample if the volume titrated was 63.0 mL. The titration occurs as per equation:

NaOH (aq) + HCl (aq) --> H2O (l) + NaCl (aq)

From the equation, No. of moles of NaOH = No. of moles of HCl
No. of mole of HCl = 0.80M x 0.0315L = 0.025 mole = No. of moles of NaOH = Molarity x Volume
So Molarity of NaOH = 0.025mole/0.063L = 0.397M ~ 0.40M

Use the formula V1M1/n1 = V2M2/n2
n1,n2 are number of moles of HCl,NaOH = 1 as per balanced equation,V1 = 31.5ml,M1=0.8M.V2=63,M2=?
M2 = V1M1/n1 x n2/v2
= 31.5 x 0.8 x 1/1 x 63 = 0.40M

From the correct balanced equation we know that 1 mole acid reacts with 1 mole base. 31.5 mL = 0.031 L. 0.031 x 0.80 = 0.0248 moles acid.

So 0.0248 moles base reacted. 63.0 mL = 0.063 L.

Molarity NaOH = moles / liter = 0.0248 / 0.063 = 0.39 M.

(.80 Mole/ L HCl)*(31.5mL)*(1L/1000ml) = (63 mL)*(1 L/1000mL)*(X mole/L NaOH)

((.80)(31.5)/1000))/((63)/1000))= 0.40 M NaOH