Calculate the pH during a titration when 10.00mL of 0.098 M HCl has reacted with 9.35 mL of 0.02 M NaOH?

Question

1944 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-31T04:02:09

#pH = 1.388#

By definition

#pH = -log[HCl]#
So,

INITIALLY
#pH = -log[.098]# = 1.009

#HCl# + #NaOH# #"----->"# #NaCL# + #H_2 O#

#color(blue)"Purpose"#: to calculate the concentration of [HCl] #color(red)"left"# after the addition of 9.35 mL of 0.02 M NaOH.

Moles of HCl:
#(.098 "mol"/L) .010 L# = .00098 mol of HCl

Moles of NaOH added
#(.02 "mol"/L) .00938 L# = .0001876 mol of NaOH

Stoichiometry molar report is HCl = NaOH

Moles of left of #HCl = .00098 - .0001878 = .0007924"#

New concentration of #HCl = "moles left" / "total volume created"#

#HCl = (.0007924 "_mol") / "(.010 + .00938) L"# = #.04095 M#

FINALLY

#pH = -log[.04095]# #= 1.388#