gchem question

[vmaggio]

---

Members do not see this ad.

What is the resulting pH when 0.005 moles of KOH are added to 0.100 L of a buffer mixture which is 0.100 M in the dihydrogen phosphate and hydrogen phosphate anions? Ka2 = 6.2 x 10-8 for H2PO4-.

A. 5.21
B. 5.61
C. 6.73
D. 7.69










can anyone walk me through this problem?


thanks

 

[DamienPro]

What is the resulting pH when 0.005 moles of KOH are added to 0.100 L of a buffer mixture which is 0.100 M in the dihydrogen phosphate and hydrogen phosphate anions? Ka2 = 6.2 x 10-8 for H2PO4-.

A. 5.21
B. 5.61
C. 6.73
D. 7.69

can anyone walk me through this problem?

thanks

H2PO4- <---> H+ + HPO42-

0.100 M 6.2 x 10-8 M 0.100 M (Buffer)
0.050 M ? 0.150 M (After OH- addition)

[H+][HPO42-]/[H2PO4-] = 6.2 x 10-8
[H+][0.150]/[0.050] = 6.2 x 10-8
[H+] = 2.067 x 10-8

pH = - log [H+] = - log [2.067 x 10-8] = 7.68

Ans: D

 

[pmantz]

What is the resulting pH when 0.005 moles of KOH are added to 0.100 L of a buffer mixture which is 0.100 M in the dihydrogen phosphate and hydrogen phosphate anions? Ka2 = 6.2 x 10-8 for H2PO4-.

A. 5.21
B. 5.61
C. 6.73
D. 7.69










can anyone walk me through this problem?


thanks

Henderson-Hauselbauch and then plug and chug....