If the Kb of a weak base is 1.5 × 10^-6, what is the pH of a 0.25 M solution of this base?
I found that I need to use the quadratic equation but I'm having difficulties because I come up with taking the square root by zero and that isn't right. Please help!
a=1, b=1.5x10^-6,
c=(1.5x10^-6(.25))=3.75x10^-7
2 answers
Figured it out, it is 11. :)
You don't need to solve a quadratic.
1.5E-6 = (x)(x)/(0.25-x)
0.25-x = essentially 0.25 since x is very small.
x = 6.12E-4 = (OH^-)
Then pOH = ? and
pH + pOH = pKw = 14.
Solve for pH. something like 11 or so.
1.5E-6 = (x)(x)/(0.25-x)
0.25-x = essentially 0.25 since x is very small.
x = 6.12E-4 = (OH^-)
Then pOH = ? and
pH + pOH = pKw = 14.
Solve for pH. something like 11 or so.
1 answer