man College CHE 168-Spg18 Grac 2/25/2018 11:55 PM 6.4/10 11 caoulator incorrectI

Question

man College CHE 168-Spg18 Grac 2/25/2018 11:55 PM 6.4/10 11 caoulator incorrectIncorectIncorredt Print -Periodic Table Question 5 of 6 Map d a Sapling Learning and K2 3.2x10- Given a diprotic acid, H2A, with two ionization constants of K4910 calculate the pH for a 0.109 M solution of NaHA. Number pH-4.88 There Is a hint available View the hint by clicking on the bottorm divider bar. Click on the divider bar again to hide the hint Close no Previous Grve Up & View Solution D check Answer Next El Exit- about us careersprivacy policy terms of use arch DOLL

Explanation / Answer

concentration of NaHA= 0.109M

NaHA---------- Na+ + HA-

HA- --------------- H+ + A-2

0.109              0          0

-x                  +x        +x

0.109-x             +x        +x

Ka1= [H+][A-2]/[HA-]

4.9x10^-4 = x*x/(0.109-x)

for solivng the equation

x= 0.00707

[H+]= 0.00707M

-log[H+]= -log(0.00707)

PH= 2.15

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