How many milliliters of 0.174 M HCl should be added to 61.8 mL of 0.0134 M potas
ID: 496234 • Letter: H
Question
How many milliliters of 0.174 M HCl should be added to 61.8 mL of 0.0134 M potassium hypobromite to give a pH of 8.43? Starting with the fully protonated species, write stepwise acid dissociation reactions for EDTA. Be sure to remove the protons in the correct order. Then starting with the fully deprotonated species, write stepwise base hydrolysis reactions for EDTA. Label each reaction with K_1 or K_b (K_a1, K_b1, K_a2 ... etc.). Calculate the pH, [H_2 L^+], [HL], [L^-] for each solution of serine when the solution is 0.0250 MH_2 L^+ 0.0750 M L^- 0.0670 M HL Find the pH of the buffer prepared by dissolving 2.67 g of glutamine chloride and 3.47 g of glutamine in 100 mL of H_2 O. What will the new pH be if 10 mL of 0.9457 M NaOH is added to the buffer? What will the new pH be if 15 mL of 0.8157 M HNO_3 is added to the buffer? Find the pH of 0.01 M H_3 Cys^2+, 0.01 M H_2 Cys^+, 0.01 M HCys, 0.01 M Cys- where Cys stands for the amino acid cysteine.Explanation / Answer
1) potasssium hypobromite = KOBr
[KOBr] = molarity x volume in Litres = 0.0134 M x 0.0618 L = 0.00083 mol
[HCl] = molarity x volume in Litres = 0.174 M x V L = 0.174 V mol
KOBr + HCl --------------> HOBr + KCl
0.00083 0.174 V 0
---------------------------------------------------------------------------------------------
0.00083 - 0.174V 0 0.174 V
Hence,
[KOBr] = 0.00083 - 0.174V
[HOBr] = 0.174 V
Given that pH = 8.43
pKa of HOBr = 8.6
From Henderson-Hasselbalch's equation,
pH = pKa + log [KOBr] /[HOBr]
8.43 = 8.6 + log ( 0.00083 - 0.174V / 0.174 V)
(0.00083 - 0.174V ) / 0.174 V = 0.676
0.00083 - 0.174V = 0.1176V
V = 0.0028 L
= 2.8 mL
V = 2.8 mL
Therefore,
volume of HCl required = 2.8 mL
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