Question

100ml of solution of HCl of 3.65 g L^-1 are mixed with 100Ml of NaOH containing 2 g L^-1 what is the pH of resulting Solution?

Answer 1

@Kryten

Answer 2

ok lets take it long way round
gama - mass concentration
gama = m / V
m= gama*V
m(HCl) = 3,65 g/dm3 * 100 * 10-3 dm3
m(HCl)= 0,365 g
n= m/M
n(HCl)= 0,365 g / 36,5 gmol-1 = 0,01 mol

m(NaOH) = 2 g/dm3 * 100*10-3 dm3
m(NaOH) = 0,2 g
n(NaOH) = 0,2 g / 40 gmol-1 = 0,005 mol

and if

n(HCl)/n(NaOH) = 1/1 = 1

from reaction

HCl + NaOH -> NaCl + H2O

than

n(HCl) that remains is

n(HCl) - n(NaOH) = 0,01 - 0,005 = 0,005

and n(HCl) = n (H+)

c(H+) = 0,005 mol / 0,2 dm3
c(H+) = 0,025 M

pH = - log [H+]
pH = 1,6