Question

When 50.0mL of 0.100 M NaOH and 50.0mL of 0.100 M HCL are mixed. what is the pH of the resultant solution? (show two chemical equations to obtain [H+] and then calculate pH)

Answer 1

Given

NaOH

V1 = 50 ml = 0.05 L

M1 = 0.1 M (mol/L)

No. of moles of NaOH = V1*M1 = 0.05 L * 0.1 (mol/L) = 0.005 moles

NaOH (aq) ------> Na+ (aq) + Cl- (aq)

so there will 0.005 moles of Na+ and OH- each

HCl

V2 = 50 ml = 0.05 L

M2 = 0.1 M (or mol/L)

No. of moles of HCl = V2* M2 = 0.05 L * 0.1 mol/L = 0.005 moles of HCl

HCl (aq) --------> H+ (aq) + Cl- (aq)

so 0.005 moles of H+ and Cl- each will be produced

all the 0.005 moles of H+ will be nuetralized by 0.005 moles of OH-

so there will be no H+ remaining

[H+] = [OH-]

taking log on both sides

-log ([H+]) = - log ([OH-])

pH = pOH

and we know that for any solution pH + pOH = 14

so sub above eqn pH + pH = 14

pH = 7