............NaOH + HX ==> NaX + H2O
begin.....M x L...M x L.....0....0
end..see below.
NaOH = moles begin.
HX = moles begin.
Reaction produces the smaller moles NaX (which will be the moles NaOH)
The difference larger-smaller = moles HX at end so NaOH will be zero at end and HX will be the moles HX-moles NaOH.
Ka = (H^+)(X^-)/(HX)
Substitute and solve for Ka.
when 15.12 ml of 0.1194 M NaOH is added to 25.00 ml of 0.1246 M HX ( an unknown weak acid), the resulting pH is 5.9. What is the Ka of the unknown acid?
1 answer
1 answer