Question about pH?
In a 1 Litre solution, 0.25 moles of acetic acid are mixed with 0.20 moles of sodium acetate. What is the pH of this solution?
Answer:
It is a buffer solution.
[CH3COOH ] = 0.25 / 1 = 0.25 M
[ CH3COO- ] = 0.20 / 1 = 0.20 M
Ka = 1.8 x 10^-5
pKa = - log Ka = 4.74
You can use the Hasselbalch's equation
pH = pKa + log [ CH3COO-] / [CH3COOH]
pH = pK + log 0.20/0.25 = 4.64
A process for extracting metal values especially zinc values from aqueous solutions of metal salts which comprises contacting the aqueous solution with an organic phase comprising a compound of the formula: ##STR1## wherein each of R.sup.1, R.sup.2, R.sup.3 and R.sup.4, independently, represents an optionally substituted hydrocarbyl or hydrocarbyloxy group or R.sup.1 and R.sup.2 together with the attached phosphorus atom and/or R.sup.3 and R.sup.4 together with the attached phosphorus atom form a 5- to 8-membered heterocyclic ring.
You have not given PKa of acetic acid
On the link below I found 4.75
So , formula of hasselbach
pH = pK +log ([salt]/[acid])
pH = pK +log 0.20/0.25 = 4.75+log 0.2/0.25=4.75+log0.8=4.65
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Answer:
It is a buffer solution.
[CH3COOH ] = 0.25 / 1 = 0.25 M
[ CH3COO- ] = 0.20 / 1 = 0.20 M
Ka = 1.8 x 10^-5
pKa = - log Ka = 4.74
You can use the Hasselbalch's equation
pH = pKa + log [ CH3COO-] / [CH3COOH]
pH = pK + log 0.20/0.25 = 4.64
A process for extracting metal values especially zinc values from aqueous solutions of metal salts which comprises contacting the aqueous solution with an organic phase comprising a compound of the formula: ##STR1## wherein each of R.sup.1, R.sup.2, R.sup.3 and R.sup.4, independently, represents an optionally substituted hydrocarbyl or hydrocarbyloxy group or R.sup.1 and R.sup.2 together with the attached phosphorus atom and/or R.sup.3 and R.sup.4 together with the attached phosphorus atom form a 5- to 8-membered heterocyclic ring.
You have not given PKa of acetic acid
On the link below I found 4.75
So , formula of hasselbach
pH = pK +log ([salt]/[acid])
pH = pK +log 0.20/0.25 = 4.75+log 0.2/0.25=4.75+log0.8=4.65
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