A 50ML solution of .35M C5H6O4 (ka 1.4 x 10^-4) is titatredwith .173 M Koh... a)
ID: 75757 • Letter: A
Question
A 50ML solution of .35M C5H6O4 (ka 1.4 x 10^-4) is titatredwith .173 M Koh...a) how many ML of KOHis required to reach equivalence point?
b) what is the PH at midpoint? for this questionwe do not divide bytwo...I believe, but i dotn knwo why...
C) what is the PH at end point...? A) .05L x .35M = .0175.... .0175/.173MKoH= .1011 L--->101.1 ML(the ans asks forML) B) SINCE THE GIVEN IS KA the PH IS JUST THE NEGATIVELOG... -log 1.4 x 10-4 = 3.85 = PH C) PH at endpoint** this is where i have alittle trouble KOH --> K+ + OH- .0175 / .035L + .1011===== .128 x2/.128= 1.4 x 10^-4 x2= 1.8 x 10^ -5 X2= .0042 -log .0042= 2.37 Poh 14-2.37= 11.62= PH can somoen tell me if i did this right?? A 50ML solution of .35M C5H6O4 (ka 1.4 x 10^-4) is titatredwith .173 M Koh...
a) how many ML of KOHis required to reach equivalence point?
b) what is the PH at midpoint? for this questionwe do not divide bytwo...I believe, but i dotn knwo why...
C) what is the PH at end point...? A) .05L x .35M = .0175.... .0175/.173MKoH= .1011 L--->101.1 ML(the ans asks forML) B) SINCE THE GIVEN IS KA the PH IS JUST THE NEGATIVELOG... -log 1.4 x 10-4 = 3.85 = PH C) PH at endpoint** this is where i have alittle trouble KOH --> K+ + OH- .0175 / .035L + .1011===== .128 x2/.128= 1.4 x 10^-4 x2= 1.8 x 10^ -5 X2= .0042 -log .0042= 2.37 Poh 14-2.37= 11.62= PH can somoen tell me if i did this right??
Explanation / Answer
a) 0.05 L x 0.35 M = 0.173M x L x L = 0.05 L x0.35 M / 0.173 M = 0.1011 L b) Mid point means pH = p Ka = - log ( 1.4 x 10^-4) =3.85 c) pH at endpoint isgreater than 7. C5H5O4- (aq) + H2O <------------> C5H6O4(aq) + OH- (aq) .0175 / .035L + .1011=====.128 C5H5O4- (aq) + H2O <------------> C5H6O4(aq) + OH- (aq) I(M) 0.128 0 0 C(M) -x +x +x E(M) 0.128-x +x +x Kb = (x)(x) / 0.128 -x ) Kb = 1.0 x 10-14 / 1.4 x 10^-4 = 7.1 x 10-11 x^2/(.128-x) = 1.0 * 10-14 / 1.4 x10^-4 x2= 0.128 M *7.1 x 10-11 x =3.02 * 10-6 M pOH = 5.5 pH = 8.48 c) pH at endpoint isgreater than 7. C5H5O4- (aq) + H2O <------------> C5H6O4(aq) + OH- (aq) .0175 / .035L + .1011=====.128 C5H5O4- (aq) + H2O <------------> C5H6O4(aq) + OH- (aq) I(M) 0.128 0 0 C(M) -x +x +x E(M) 0.128-x +x +x Kb = (x)(x) / 0.128 -x ) Kb = 1.0 x 10-14 / 1.4 x 10^-4 = 7.1 x 10-11 x^2/(.128-x) = 1.0 * 10-14 / 1.4 x10^-4 x2= 0.128 M *7.1 x 10-11 x =3.02 * 10-6 M pOH = 5.5 pH = 8.48 C5H5O4- (aq) + H2O <------------> C5H6O4(aq) + OH- (aq) .0175 / .035L + .1011=====.128 C5H5O4- (aq) + H2O <------------> C5H6O4(aq) + OH- (aq) I(M) 0.128 0 0 C(M) -x +x +x E(M) 0.128-x +x +x Kb = (x)(x) / 0.128 -x ) Kb = 1.0 x 10-14 / 1.4 x 10^-4 = 7.1 x 10-11 x^2/(.128-x) = 1.0 * 10-14 / 1.4 x10^-4 x2= 0.128 M *7.1 x 10-11 x =3.02 * 10-6 M pOH = 5.5 pH = 8.48Related Questions
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