1) Calculate the volume of 0.100 M HCl that is required to react with 50.0 mL of
ID: 851740 • Letter: 1
Question
1) Calculate the volume of 0.100 M HCl that is required to react with 50.0 mL of 0.100 M ammonia to form a mixture that has the desired pH. Hint: think of this as a titration problem. The desired pH is 9.00.
2) What is the pH of a solution prepared by adding 25.0 mL of 0.10 M acetic acid (Ka= 1.8*10^-5) and 20.0 mL of 0.10 M sodium acetate? This one I already know the answer to.
What is the pH after the addition of 1.0 mL of 0.10 M HCl to 20 mL of the solution above. (this is the one I have trouble with) (1 and 2 are not relate)
Explanation / Answer
1) Let Xml of Hcl is used , hence
HCl= x*0.1 = 0.x mmoles , NH3 = 50*0.1 = 5 mmoles
NH3 + HCl => NH4Cl
5 0.x 0
5-0.x 0.x
The ammonium ions are weakly acidic, and this equilibrium is set up whenever they are in solution in water
NH4+ => NH3 + H+
Ka= [NH3][H+]/[NH4+] , Ka for the ammonium ion is 5.62 x 10-10 mol dm-3. and desired pH is 9 that is H+= 10^-9
so putting the values
5.62 *10^-10 = (5-0.x)*(10^-9)/ 0.x
x = 32 ml ( of HCl )
2) acetic acid = 25 * 0.1 = 2.5 / 45 M
sodium acetate =20* 0.1 = 2/ 45 M
Both these together form an acidic buffer and the pH is given by
Henderson-Hasselbalch equation : pH= pKa + log ( A-/HA)
ka= 1.8*10^-5 ; pKa = 4.74
pH = 4.74 + log( 2 / 2.5 ) = 4.52
3) Acetic acid = 0.0556 M , Sodium Acetate = 0.0444 M
addition of 1.0 mL of 0.10 M HCl to 20 mL of the solution ;
we have Acetic acid = 0.0556* 20 + 0.1(H+ from HCl) = 1.212 mmoles
and Sodium acetate = 0.0444*20 - 0.1(H+ from HCl) = 0.788 moles
Ka= H+* Ch3CooNa / Ch3CooH
1.8*10^-5 = H+ * 0.788/ 1.212 ; H+ = 2.76*10^-5 ; pH=4.55
Thanks :)
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