1. The document provides instructions to calculate the pH of several solutions. It first asks to calculate the pH of 0.15 M NaOH, which is found to be 13.18. It then asks to calculate the pH of 0.15 M NH3 using its dissociation constant, which is found to be 11.21 through solving a quadratic equation. The document does not provide answers for questions 3-5.

1. 1. Calculate the pH of 0.15 M NaOH
2. Determine pH for a solution that is 0.15 M in NH 3 . ( K b = 1.76 x 10 -5
)
3. ?Determine the pH of a solution that is 0.15 M in NH 4 Cl at 25°C and determine if it is
acidic or basic. ( K b (NH 3 ) = 1.76 x 10 -5
)?
4.Calculate the pH of a buffer solution that is 0.15 M NH 3 ( K b = 1.76 x 10 -5
) and 0.15 M NH 4
Cl.
5. A 1.00-L buffer solution is initially 0.15 M in NH 3 ( K b = 1.76 x 10 -5
) and 0.15 M in NH 4
Cl. Find the pH after 0.010 mol of solid NaOH are added.
Solution
1)
[OH-] = [NaOH] = 0.15 M
use:
pOH = -log [OH-]
= -log (0.15)
= 0.824
use:
PH = 14 - pOH
= 14 - 0.824
= 13.18
Answer: 13.18

2. 2)
NH3 dissociates as:
NH3 +H2O -----> NH4+ + OH-
0.15 0 0
0.15-x x x
Kb = [NH4+][OH-]/[NH3]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((1.76*10^-5)*0.15) = 1.625*10^-3
since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Kb = x*x/(c-x)
1.76*10^-5 = x^2/(0.15-x)
2.64*10^-6 - 1.76*10^-5 *x = x^2
x^2 + 1.76*10^-5 *x-2.64*10^-6 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 1.76*10^-5
c = -2.64*10^-6
Roots can be found by

3. x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a
b^2-4*a*c = 1.056*10^-5
roots are :
x = 1.616*10^-3 and x = -1.634*10^-3
since x can't be negative, the possible value of x is
x = 1.616*10^-3
So, [OH-] = x = 1.616*10^-3 M
use:
pOH = -log [OH-]
= -log (1.616*10^-3)
= 2.7916
use:
PH = 14 - pOH
= 14 - 2.7916
= 11.21
Answer: 11.21
Only 1 question at a time please