ex . So, after the addition of Ha, the newpH will be-log(2.01×10-2)=1.86. The ex
ID: 706717 • Letter: E
Question
ex . So, after the addition of Ha, the newpH will be-log(2.01×10-2)=1.86. The expected change in pH will be 1.86-2.50--0.64 pH units. Additional Questions (for the finished lab report) 1· For each of these desired pH values, choose a weak acid congugate base pair from your textbook that could be used to prepare a buffer solution with that pH. Calculate the desired [A VIHA] ratio in each case. (a) pH = 6.00 (b) pH- 8.00(e) pH 11.00 2. 500 mL of a buffer solution contains 0.050 mol NaHSO3 and 0.031 mol Na2sO;3 (a) What is the pH of the solution? (b) Write the net ionic equation for the reaction that occurs when NaOH is added to this buffer. (c) Calculate the new pH after 10. mL of 1.0 M NaOH is added to the buffer solution. (d) Calculate the new pH after 10. mL of 1.0 M NaOH is added to 500. mL of pure water. (e) Explain why the pH of the water changed so much as compared to the pH of the buffer.Explanation / Answer
ans)
1)
Here we can using the weak acids according to their ka values
here we can use three weak acids
1.carbonic acid
2.dihydrogen phospate(monobasic)
3.monohydrogen phospate
a) at pH = 6.00
The corresponding Ka of carbonic acid = 4.6 x10^-7
pKa = -log Ka = -log(4.6 x10^-7) = 6.34
the buffer = H2CO3 + HCO3-
For acidic buffer
Henderson-Hasselbalch equation
pH = pKa + log[salt/acid]
6 = 6.34 + log[HCO3-/H2CO3]
log[HCO3-/H2CO3] = -0.34
[HCO3-/H2CO3] = 0.457
b)
at
pH = 8.00
the corresponding Ka of dihydrogen phosphate = 6.2 x10^-8
pKa = -log Ka = -log(6.2 x10^-8) = 7.21
here the buffer = H2PO4- + HPO4-2
pH = pKa + log[salt/acid]
8 = 7.21 + log[HPO4-2/H2PO4-]
log[HPO4-2/H2PO4-] = 0.79
[HPO4-2/H2PO4-] = 6.165
c)
at pH = 11.00
the corresponding Ka of monohydrogen phosphate = 4.8 x10^-13
pKa = -log Ka = -log( 4.8 x10^-13) = 12.32
here the buffer = HPO4-2 + PO4-3
pH = pKa + log[salt/acid]
11 =12.32 + log[PO4-3/HPO4-2]
log[PO4-3/HPO4-2] = -1.32
[PO4-3/HPO4-2] =0.0478
2)
a)
Given
the moles in ]NaHCO3=O.O50 mol
the moles in Na2SO3=0.031 M0L
in this HSO3- Pka value IS = 7.21
pH = pKa + log [salt/acid]
= 7.21 + log (0.031/0.050)
= 7.002
The PH value is=7.002
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