Method 1: Convert [OH-] to [H3O+] first, then insert [H3O+] directly into the pH equation. Type your numeric answer and submit 1e1 Answered - Incorrect 1 attempt left Q15.37 • Unanswered. Method 2: Convert [OH-] to pOH, then convert pOH to pH. Type your numeric answer and submit × You are incorrect Resubmit Open

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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Step 5: Once [OH-] at equilibrium is known, we can use one of two ways to determine the pH of the weak
base, as covered in Section 15.4.
Q15.36
Method 1: Convert [OH-] to [H3O+] first, then insert [H3O+] directly into the pH equation.
1e1
• Answered.
Type your numeric answer and submit
Answered - Incorrect. 1 attempt left
Q15.37
• Unanswered.
Method 2: Convert [OH-] to pOH, then convert pOH to pH.
Type your numeric answer and submit
X
You are incorrect
Resubmit
7
Open in Re
Transcribed Image Text:Step 5: Once [OH-] at equilibrium is known, we can use one of two ways to determine the pH of the weak base, as covered in Section 15.4. Q15.36 Method 1: Convert [OH-] to [H3O+] first, then insert [H3O+] directly into the pH equation. 1e1 • Answered. Type your numeric answer and submit Answered - Incorrect. 1 attempt left Q15.37 • Unanswered. Method 2: Convert [OH-] to pOH, then convert pOH to pH. Type your numeric answer and submit X You are incorrect Resubmit 7 Open in Re
Step 4: Plug the equilibrium expressions into the equation for K and solve for x.
[OH Jeq[NH4*Jeq
Kb =
[NH3Jeg
=
x²
2
1.2 - X
Equation 15.30
x²
1.2-X
+²
Because K₁ <10-5, the method of successive approximations is worth a try:
=
x²
2
[NH3], -X
1.8 x 10-5
≈1.8x10-5
1.2
x=4.6 x 10-³ M = [OH Jeg
-3
Equation 15.31
0.05 x 1.2 M = 0.06 M > 4.6 x 10-³ M
=
Equation 15.32
1.8 x 10-5
Remember, when K is on the borderline value of ≈10-5, it's a good habit to check and see if x < 5% of the
initial concentration of the original molecule.
.. assumption is valid
K
7
Open in
Transcribed Image Text:Step 4: Plug the equilibrium expressions into the equation for K and solve for x. [OH Jeq[NH4*Jeq Kb = [NH3Jeg = x² 2 1.2 - X Equation 15.30 x² 1.2-X +² Because K₁ <10-5, the method of successive approximations is worth a try: = x² 2 [NH3], -X 1.8 x 10-5 ≈1.8x10-5 1.2 x=4.6 x 10-³ M = [OH Jeg -3 Equation 15.31 0.05 x 1.2 M = 0.06 M > 4.6 x 10-³ M = Equation 15.32 1.8 x 10-5 Remember, when K is on the borderline value of ≈10-5, it's a good habit to check and see if x < 5% of the initial concentration of the original molecule. .. assumption is valid K 7 Open in
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