ammonium ion - ammonia C. Some Properties of Buffers Buffer system selected. NH4-NH3 HB is 1. pH of buffer 2. pH of diluted buffer 156 9.54 9.48 NH4CI CO.10 M) M pK, (by Eq. 4b) pk, [H*]. [H*]. ? pH after addition of excess NaOH Experiment 19 pH Measurements-Buffers and Their Properties 3. pH of buffer in which [HB]/[B] = 0.10 M (name the acid) ? 9.50. 10.54 11.83 .pk₁

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Chapter1: Chemical Foundations
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Calculate H+ and pka (given is pH) refer to the picture below. Please include your solution. PLEASE ANSWER QUESTIONS 1-3 and use the 2nd picture for the formula that must be used
ammonium ion - ammonia
C. Some Properties of Buffers
Buffer system selected. NH4-NH3 HB is
1. pH of buffer
2. pH of diluted buffer
156
9.54
9.48
[H*].
[H*]
NH4Cl CO.10 M)
M pK, (by Eq. 4b)
M
pk,
?
Experiment 19 pH Measurements-Buffers and Their Properties
3. pH of buffer in which [HB]/[B] = 0.10
pH after addition of excess NaOH
(name the acid)
?
?
9.50.
10.54
11.83
pK₁
Transcribed Image Text:ammonium ion - ammonia C. Some Properties of Buffers Buffer system selected. NH4-NH3 HB is 1. pH of buffer 2. pH of diluted buffer 156 9.54 9.48 [H*]. [H*] NH4Cl CO.10 M) M pK, (by Eq. 4b) M pk, ? Experiment 19 pH Measurements-Buffers and Their Properties 3. pH of buffer in which [HB]/[B] = 0.10 pH after addition of excess NaOH (name the acid) ? ? 9.50. 10.54 11.83 pK₁
a strong
The pH of such a buffer is established by the relative concentrations of HB and B in the solution. If we
manipulate Equation 4, we can solve for the concentration of H+ ion:
[H] = K₂ X [HB]/[B]
(4a)
This equation is often further modified when working with buffers, by taking the negative logarithm to base
10 of both sides:
pH pK,+ log [B]/[HB]
(4b)
Transcribed Image Text:a strong The pH of such a buffer is established by the relative concentrations of HB and B in the solution. If we manipulate Equation 4, we can solve for the concentration of H+ ion: [H] = K₂ X [HB]/[B] (4a) This equation is often further modified when working with buffers, by taking the negative logarithm to base 10 of both sides: pH pK,+ log [B]/[HB] (4b)
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