Question

Calculate the pH of a 0.025M solution of ammonia Kb =1.75x10^-5

Answer 1

Calculate the pH of 0.025M solution ammonia, Kb =1.75x10^-5

Answer 2

Bot, I believe you made a math error in the quadratic formula. The equation you have of x^2 + 1.75 x 10^-5x - 4.375 x 10^-7 = 0 is correct but I have (OH^-) = approximately 0.00067 M which makes the pH = approximately 10.8. Check it out please.

Answer 3

Bot, you have an error in the quadratic formula. You are correct to this point.
x = (-1.75 x 10^-5 ± √[(1.75 x 10^-5)^2 - 4(1)(-4.375 x 10^-7)]) / (2(1)). However, in the next step it appears you have multiplied by an extra 4.375 x 10^-7; i.e. in the quadratic formula it is -4ac. You have multiplied by 4ac*c and the extra c is not needed. Check it out please.

x ≈ (-1.75 x 10^-5 ± √[3.0625 x 10^-10 + 1.75 x 10^-6 x 4.375 x 10^-7]) / (2)

In addition, note that plugging your answer into Kb does not give Kb and it should if your value is correct. Plugging my value into Kb satisfies the equation.

Answer 4

Thank you.