Asked by Bradely
                If 25.00 mL of 0.20 mol/L HCO2H(aq) is titrated with 0.20 mol/L NaOH(aq) (the titrant), determine the pH
a) befor titration began
b)after 10.00 mL of NaOH has been added
c) at the equivalence point
            
        a) befor titration began
b)after 10.00 mL of NaOH has been added
c) at the equivalence point
Answers
                    Answered by
            DrBob222
            
    a) before titration begins. That is a 0.2 M solution of HCOOH.
HCOOH ==> H^+ + HCOO^-
(HCOOH) = 0.2-x
(H^+) = x
(HCOO^-) = x
Ka = (H^+)(HCOO^-)/(HCOOH)
Plug the above variables into Ka and solve for (H^+), then convert to pH.
b) after 10.00 mL NaOH has been added.
HCOOH + NaOH ==> HCOONa + H2O
mols HCOOH initially = M x L = ??
mols NaOH added = M x L = ??
mols HCOONa formed = ??
pH = pKa + log[(base)/(acid)]
c)at the equivalence point. We have the salt HCOONa + H2O at the equivalence point.
HCOO^- + HOH ==> HCOOH + OH^-
Kb = Kw/Ka = (HCOOH)(OH^-)/(HCOO^-)
(HCOOH) = x
(OH^-) = x
(HCOO^-) = 0.2 - x
Plug into Kb and solve for x = (OH^-), then convert to pH.
Post your work if you get stuck.
Check my thinking. Check my work. Check my arithmetic.
                    Answered by
            Sarah
            
    a) Ka = (H^+)(HCOO^-)/(HCOOH)
= (x)(x)/(0.2-x)
= (x^2) / 0.2 - x
Next step please...I Hate these problems :P:)
    
= (x)(x)/(0.2-x)
= (x^2) / 0.2 - x
Next step please...I Hate these problems :P:)
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