pH of resulting solution (1 Viewer)

[Dubble-u25]

In an experiment, a student mixed 15.0mL of 0.030mol L^-1 HCl with 20.0mL of 0.010mol L^-1 Ba(OH)₂. Calculate the pH of the resulting solution.
The answer is 6.76 but I got 2.845.
As you can see I'm way off. Please help

 

[Dylanamali]

WRITE THE BALANCED EQUATION.

Ba(OH)2(aq) + 2HCl(aq) -> BaCl2(aq) + H2O(l)

n of HCl initial = C x V = 0.030 x 0.015 = 0.00045 moles

n of Ba(OH)2 = 0.010 x 0.020 = 0.0002 moles

from eqn) -> n Ba(OH)2 : n HCl = 1 : 2
therefore, n of HCl reacted = 2 x n Ba(OH)2 = 2 x 0.0002 = 0.0004 moles

n HCl excess = n HCl initial - n HCl reacted = 0.00045 - 0.0004 = 0.00005 moles
C HCl = n/V = 0.00005/0.035 = 0.00143 M

HCl -> H+ + Cl- [monoprotic]
therefore [HCl] = [H+] = 0.00143M
pH = -log[H+] = -log(0.00143) = 2.845

 

[Dubble-u25]

WRITE THE BALANCED EQUATION.

Ba(OH)2(aq) + 2HCl(aq) -> BaCl2(aq) + H2O(l)

n of HCl initial = C x V = 0.030 x 0.015 = 0.00045 moles

n of Ba(OH)2 = 0.010 x 0.020 = 0.0002 moles

from eqn) -> n Ba(OH)2 : n HCl = 1 : 2
therefore, n of HCl reacted = 2 x n Ba(OH)2 = 2 x 0.0002 = 0.0004 moles

n HCl excess = n HCl initial - n HCl reacted = 0.00045 - 0.0004 = 0.00005 moles
C HCl = n/V = 0.00005/0.035 = 0.00143 M

HCl -> H+ + Cl- [monoprotic]
therefore [HCl] = [H+] = 0.00143M
pH = -log[H+] = -log(0.00143) = 2.845

Cool so I'm right? Thanks

 

[Antadhur123]

No you're not