The pH of the buffer is 4.523 .
Given,
A buffer is prepared by adding 300 ml of 2M NaOH to 500 ml of 2M CH3COOH .
Thus , the volume of NaOH =300ml
volume of CH3COOH = 500ml
molarity of NaOH = 2M
Molarity of CH3COOH =2M
because of adding the two compound ,
the total volume =( 300+500)ml =800ml
We know ,
number of moles = molarity ×volume
number of moles of NaOH = 300ml×2M = 600mol
number of moles of CH3COOH = 500ml×2M = 1000ml
Due to mixing of NaOH and CH3COOH , there is a change in volume and concentration .
We know ,
molarity = number of moles /total volume
concentration of NaOH = 600mol/800ml = 0.75M
concentration of CH3COOH = 1000mol/800ml =1.25M
We know ,
pH = pKa + log {[base ]/[acid]}
pH = -log(1.8×10^-5) + log{[0.75]/[1.25]}
pH = 4.744 + (-0.221)
pH = 4.523
Hence , the pH of the buffer is 4.523 .
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