What is the pH of a 0.350 M solution of lithium citrate? (Ka Citric Acid = 7.4 x 10-4)

Question

I'm struggling to answer this one. Would appreciate a simple explanation if possible.

Answer 1

They want you to find pH of lithium citrate, but they gave you the K_a for citric acid, so now you have to think:

If they gave you the K_a of something you can't solve directly, then you have to adapt to what is given.

Given the K_a of citric acid, then lithium citrate must be a base (technically a salt). So you can use the K_a to find the K_b by using the following equation:

K_a x K_b = K_w Kw = 10^-14

rearrange this equation you get:

K_b = K_w/K_a

= 10^-14/(7.4 x 10^-4) = 1.35 x 10^-11

Now since we are dealing with K_b instead of finding the [H₃O⁺] concentration, you need to find the [OH^-] concentration by using the ICE table. We know the concentration of the salt is 0.350 M

Chemical Reaction: salt + water ---> acid + base

I 0.350 0 0

C -x +x +x

E 0.350-x x x

K_b = [OH^-][H₃O⁺]/[Base]

= (x)(x)/(0.350 - x)

= x²/(0.0350 - x) here we can assume that x is small and it can be ignored, but for future references, always do the approximation to be sure. Now we just need to solve for x

x² = K_b x 0.350 to get rid of the square root you have to square both sides√

√x² = √(1.35 x 10^-11)(0.350)

x = 2.17 x 10^-6 this is the [OH^-]

pOH = -log[OH^-]

= -log(2.17 x 10^-6) this is pOH not pH. Use the following equation to solve for pH

= 5.66

pH + pOH = 14

pH = 14 - pOH

= 14 - 5.66

= 8.34