1)What is the pH of a 0.175 mol/L solution of sodium acetate, CH3COONa, at 25oC?
ID: 748008 • Letter: 1
Question
1)What is the pH of a 0.175 mol/L solution of sodium acetate, CH3COONa, at 25oC? (pKa=4.74 at 25oC for CH3COOH.) Give your answer accurate to two decimal places. 2)How many moles of NaOH must be added to 0.550 L of 0.191 mol/L HC3H5O2(aq) to obtain a solution with pH=4.7? Assume the temperature is 25oC. (For HC3H5O2, pKa=4.89 at 25oC). Give your answer accurate to three significant figures.Explanation / Answer
Sodium acetate: CH3-COO-Na This is a salt of a weak acid (acetic acid) and a strong base (NaOH). This salt dissolves in water: CH3COONa=(CH3COO)- + (Na)+ The acetate ion reacts with water but the Na+ does not: (CH3COO)- + H2O = CH3COOH + OH- This means that your solution will be basic. This salt is a strong electrolite so it dissolves completley. The volume is irrelevant here since you have the contrencation. C(salt)=1.0159mol/dm3 since it dissolves completley, you will have the acetate ion in the same concentration.--> C (CH3COO-)=1.0159mol/dm3 There is a thing called the hidrolisis constant that looks like this: K(hidrolisis)=K(water)/K(acid) or K(hidrolisis)=K(water)/K (base). Where K(hidrolisis) is the hidrolisis constant of the ion that came from the salt, in this case it is the acetate ion. (If our salt was ammonium- chloride=NH4Cl then K (hidrolisis) would be the constant of the [NH4+] ion). K(acid) is the dissociation constant of the acid which the salt turns into during reacting with water: in this case it is the acetic acid and this is what we will need.(Again if we had NH4Cl for the salt, then we wont have a K(acid) because (NH4+) + H2O= NH3 + (H3O+) so we'll have a base formed thus we will have a K(base) for NH3). K(base): see above. So we'll need the K(acid) for CH3COOH, which usually is: Ks (CH3COOH)=1.86*10^-5 Since in this case K (hidrolisis)=K(water)/K(acid) and K(water)=10^-14--> K (hidrolisis)=5.3763*10^-10 Now that you have the K (hidrolisis) you can treat your solution as a weak base which has a base dissociation constant(Kb) of 5.3763*10^-10 and a concentration(Cb) of 1.0159mol/dm3 There are many rules that I wont explain in detail here but in weak bases if Kb/Cb is less than 10^-3 then you have to use the following equation: Kb=[OH-]^2/Cb So 5.3763*10^-10=X^2/1.0159 -- > X=[OH-]=2.337*10^-5 --> pOH=4.6313 --> pH=9.3687 I know it is difficult and long but thats the way with hidrolising salts. If something is not clear drop me a message and I'll try to explain further. Hope I helped!
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