A 35 ml solution of .241 M Hcl is titrated with .127M ofNAOH... a) how many ML o

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        = 8.48

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