# Question #672a6

##### 1 Answer

You have to look up the

I'll use the

#K_w = K_aK_b# .

Therefore, you can get the

#K_a = 10^(-14)/(1.3 xx 10^(-6)) = 7.69 xx 10^(-9)#

The reaction itself can be denoted in general:

#"BH"^(+)(aq) " "" "+" "" " "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "B"(aq)#

#"I"" ""0.10 M"" "" "" "" "" "" "-" "" "" ""0 M"" "" "" "" ""0 M"#

#"C"" "-x" "" "" "" "" "" "" "-" "" "+x" "" "" "" "+x#

#"E"" "(0.10 - x)"M"" "" "" "" "-" "" ""x M"" "" "" "" ""x M"#

And its

#K_a = x^2/(0.10 - x)#

However, since

#K_a ~~ x^2/(0.10)#

Therefore, we can solve for the equilibrium concentration of

#["H"_3"O"^(+)] = x ~~ sqrt(0.10K_a)#

#= 2.77 xx 10^(-5) "M"#

Hence, the

#color(blue)("pH") = -log["H"_3"O"^(+)]#

#= color(blue)(4.56)#

And another way to check whether the small

#%"dissoc". = x/(["soln"])#

#= (2.77 xx 10^(-5) "M")/("0.10 M") xx 100% = 0.028% "<<" 5%#

So this approximation works extremely well. You would get pretty much the same answer if you chose to solve the quadratic equation in full.