A 35 ml solution of .241 M Hcl is titrated with .127M ofNAOH... a) how many ML o
ID: 75741 • Letter: A
Question
A 35 ml solution of .241 M Hcl is titrated with .127M ofNAOH...a) how many ML of NAOH is required to reach equivalence point?
b) what is the PH at midpoint? for this question i know we divideby 2...
C) what is the PH at end point...?
v/s
A 50ML solution of .35M C5H6O4 (ka 1.4 x 10^-4) is titatred with.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...?
Part C seems to be the easiest to me, i am having trouble doingpart A and B
can someone help?
Explanation / Answer
a) 0.035 L * 0.241 M = 0.127 M *x L x = 0.035 L * 0.241 M / 0.127M =0.0664 L b) HCl + NaOH --------------> NaCl +H2O 0.035 L * 0.241M 0.0332 L *0.127 M [ H+] = 0.035 L * 0.241 M - 0.0332 L *0.127 M / 0.035 L + 0.0332 L = 0.0681 M pH = 1.2 c) At the equivalence point pH = 7 a) ) .05L x .35M = .0175M * x L .0175/.173MKOH = .1011 L = 101.1 mL b) At haf way poiny ( mid point) pH = pKa =-log 1.4 x 10^-4 =3.85 C) pH at endpoint is greater 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 = 8.48Related Questions
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