Can some please explain this answer to me please i am soo lost especially [B]=1.
ID: 946052 • Letter: C
Question
Can some please explain this answer to me please i am soo lost especially [B]=1.82[A] Let's Practice 3: Answer the Questions How many moles of sodium acetate and acetic acid are there in1 liter of a 0.1 M acetate buffer at pH 5.0. (Acetic acid has a pK of 4.74). pH = pK, + log ([Base]/[Acid]) 5.0 -log (1.8 x 10-5) +log [Bl/IA] log[B]/[A] = 5.0-4.74 [B]/[A] = 100.26 The concentration always includes both the acid and the base components of the buffer [A] =[CH,COOH] = 0.1-0.064-0.036 M [B] = 1.82[A] [B] = 1.82(0.1-(B)) [B] = 0.182. 1.82[B] 2.82[B] = 0.182 [B] = [CH3COO] = 0.064 M 1 liter of solution: A] = 0.036 moles [B] = 0.064 molesExplanation / Answer
Hi, the problem is very clear it is done excellently. i will try to explain my own way
acid = A
base = B
total buffer moles = molarity x volume
= 0.1 x 1
= 0.1
total buffer mean it is the combination of acid + base. so
A + B = 0.1 -------------------------> 1
now come to the henderson's equation : it is
pH = pKa + log [base / acid]
pH = 5.0 is given
pKa = 4.74
substitute these in above pH equation.
5.0 = 4.74 + log [B/A]
log [B/A] = 5.0 -4.74
log [B/A] = 0.26
[B/A] = 10^0.26
B / A = 1.82
B = 1.82 A ----------------------> 2
substitute equation 2 in eqation 1
A + B = 0.1
A + 1.82 A = 0.1
2.82 A = 0.1
A = 0.1 / 2.82
A = 0.036
we got A value . substitute this value in eqaution 2
B = 1.82 A
= 1.82 x 0.036
= 0.064
so finally .
moles of A = 0.036 and B = 0.064
total moles should be 0.1 you can add A + B for checking