Q1.The ammonia/ammonium buffer system is used to optimize polymerase chain react
ID: 1019004 • Letter: Q
Question
Q1.The ammonia/ammonium buffer system is used to optimize polymerase chain reactions (PCR) used in DNA studies. The equilibrium for this buffer can be written as NH4+(aq)H+(aq)+NH3(aq) Calculate the pH of a buffer that contains 0.060 M ammonium chloride and 0.090 M ammonia. The Ka of ammonium is 5.6×1010.
Q2.Water ionizes to a very small degree into hydronium ions, H3O+, and hydroxide ions, OH:
The very small value of the equilibrium constant, Kw, should give you an appreciation for how few water molecules actually ionize in pure water.
In neutral solutions, [H3O+]=[OH].
In basic solutions, [H3O+]<[OH].
In acidic solutions, [H3O+]>[OH].
But in all aqueous solutions, the product of the hydronium and hydroxide concentrations is equal to Kw. Thus, Kw allows you to calculate [H3O+]from [OH], or vice versa.
Part A
8.00×103 mol of HBr are dissolved in water to make 16.0 L of solution. What is the concentration of hydroxide ions, [OH], in this solution?
Part B
2.00 g of NaOH are dissolved in water to make 4.00 L of solution. What is the concentration of hydronium ions, [H3O+], in this solution?
2H2O(l)H3O+(aq)+OH(aq) Kw=[H3O+][OH]=1.00×1014Explanation / Answer
Q1.The ammonia/ammonium buffer system is used to optimize polymerase chain reactions (PCR) used in DNA studies. The equilibrium for this buffer can be written as NH4+(aq)H+(aq)+NH3(aq) Calculate the pH of a buffer that contains 0.060 M ammonium chloride and 0.090 M ammonia. The Ka of ammonium is 5.6×1010.
pKa of ammonium is
pKa = - log Ka
pKa = - log 5.6 x 10-10
pKa = 9.25
pH = pKa + log base/acid
pH = 9.25 + log 0.090/0.060
pH = 9.25 + 0.176
pH = 9.426
2)
Part A
8.00×103 mol of HBr are dissolved in water to make 16.0 L of solution. What is the concentration of hydroxide ions, [OH], in this solution?
HBr is a stron acid so [H3O+] = 8.0x 10-3mol/16 L = 0.0005 M
[H3O+] x [OH-] = 10-14
0.0005 x [OH-] = 10-14
[OH-] = 10-14/0.0005
[OH-] = 2 x 10-11 M
Part B
2.00 g of NaOH are dissolved in water to make 4.00 L of solution. What is the concentration of hydronium ions, [H3O+], in this solution?
2 g of NaOH is 2/40 = 0.05 moles in 4 L is 0.05/4 = 0.0125 M
since NaOH is a strong base [OH-] = 0.0125 M
[H3O+] x [OH-] = 10-14
[H3O+] x 0.0125 = 10-14
[H3O+] = 10-14/0.0125
[H3O+] = 8 x 10-13 M
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