Calculate the ph of the solution made by adding 0.50 mol of hobr and 0.30 mol of kobr to 1.00 l of water. the value of ka for hobr is 2.0×10? 9.

To fully understand the problem, we use the ICE table to identify the concentration of the species. We calculate as follows:

Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr] 

HOBr  = 0.50 M 
KOBr  = 0.30 M = OBr-

     HOBr + H2O <-> H+ + OBr- 
I     0.50        -              0          0.30 
C       -x                        x             x 

E(0.50-x)                    x       (0.30+x) 

Assuming that the value of  x is small as compared to 0.30 and 0.50 

Ka = 2.0 x 10^-9 = x (0.30) / 0.50) 

x = 3.33 x 10^-9 = H+
pH = 8.48

It would be true sodium reacts with water in a violent way