SHOW WORK Using a 0.30 M phosphate buffer with a pH of 7.7, you add 0.72...

SHOW WORK

Using a 0.30 M phosphate buffer with a pH of 7.7, you add 0.72 mL of 0.52 M HCl to 45 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)

Using a 0.30 M phosphate buffer with a pH of 7.7, you add 0.72 mL of 0.52 M NaOH to 45 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)

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Expert Solution

pH = pka + log(salt/acid)

[buffer] = [salt]+[acid]= 0.3 M

so that , let us take [acid] = x

[salt] = 0.3-x

pH = 7.7 , pka = 7.21

7.7 = 7.21+log((0.3-x)/x)

x = 0.0733 M

[acid] = x = 0.0733 M

[salt] = 0.3-0.0733 = 0.2267 M

after addition of HCl to 45 ml buffer

no of mole of salt present = 45*0.2267/1000 = 0.01

no of mole of acid present = 45*0.0733*45/1000 = 0.0033

pH = pka + log(salt-HCl/acid+HCl)

no of mole of HCl = 0.72*0.52/1000 = 0.0003744

pH = 7.21+log((0.01-0.0003744)/(0.0033+0.0003744 ))

= 7.6

after addition of NaOH

pH = pka + log(salt+NaOH/acid-NaOH)

no of mole of salt present = 45*0.2267/1000 = 0.01

no of mole of acid present = 45*0.0733*45/1000 = 0.0033

no of mole of NaOH = 0.72*0.52/1000 = 0.0003744

pH = 7.21+log((0.01+0.0003744)/(0.0033-0.0003744 ))

= 7.76

= 7.8

queen_honey_blossom answered 1 year ago

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