1-What is the pH of a buffer that consists of 0.341 M HF and 0.439 M KF? K a of
ID: 796765 • Letter: 1
Question
1-What is the pH of a buffer that consists of 0.341 M HF and 0.439 M KF? Ka of HF is 6.8 x 10-4
2-The pH of 0.50 M HClO is 3.91. Calculate the change in pH when 1.06 g of NaClO is added to 33 mL of 0.50 M HClO. Ignore any changes in volume. The Ka value for HClO is 3.0 x 10-8.
3--Calculate the pOH of a solution that results from mixing 14 mL of 0.12 M dimethylamine ((CH3)2NH) with 41 mL of 0.11 M (CH3)2NH2Cl. The Kb value for (CH3)2NH is 5.4 x 10-4.
4-A buffer consists of 0.34 M KHCO3 and 0.29 M K2CO3. Given that the K values for H2CO3 are, Ka1 = 4.5 x 10-7 and Ka2 = 4.7 x 10-11, calculate the pH for this buffer.
Explanation / Answer
pH = pka + log [KF]/[HF] , pKa = -log Ka = -log ( 6.8 x10^-4) = 3.1675
pH = 3.1675 + log ( 0.439/0.341) = 3.28
2) NaClO moles = 1.06/74.44 = 0.01424 , [NaOCl] = ( 0.01424 x1000/33) = 0.4315
pH = pka + log [NaOCl]/[HOCl] , pka = -log ( 3x10^-8) = 7.523
pH = 7.523 + log ( 0.4315/0.5) 7.46
pH change = 7.46 -3.91 = 3.55
3) pOH = pkb + log [(CH3)2NH2Cl]/[(CH3)2NH]
total vol = 14+41 = 55 ml = 0.055 L , pkb = -log ( 5.4 x10^-4) = 3.27
hence new Molarities are [(CH3)2NH] = ( 0.12 x (0.014/0.055) = 0.0305 ( M1V1 = M2V2)
[(CH3)2NH2Cl] = ( 0.11x0.041/0.055) = 0.082
pOH = 3.27 + log ( 0.082/0.0305) = 3.7 ,
4) pH = pka2 + log [K2CO3]/[KHCO3] , pka2 = -log ( 4.7 x10^-11) = 10.33
pH = 10.33 + log ( 0.29/0.34) = 10.26
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