Question

What volume of 6.00 m naoh would be required to increase the ph to 4.93?

Answer 1

according to this formula:
PH= Pka + ㏒[A^-]/[HA]
when we have the value of PH=4.93 & Pka (missing in your question) = 4.76 &
no.of mole of acetate = 40 mmol & no.of mole of acetic acid= 60
so by substitution:
4.93 = 4.76 + ㏒((40+X)/(60-X))
0.17 = ㏒ ((40+X)/(60-X))
∴X= 19.66 mmol = 0.019 mol
finally we can get the volume from this formula of the molarity:
molarity = number of moles / volume 
      6      = 0.019 / V
∴V= 0.0032 mL = 3.2 L

Answer 2

The Volume of  NaOH is mathematically given as

V=3.2 L

Volume of  NaOH

Generally the formula for PH is

Ph= Pka + log[A^-]/[HA]

4.93 = 4.76 + log((40+X)/(60-X))

0.17 = log ((40+X)/(60-X))

Therefore

X= 19.66 mmol = 0.019 mol

Generally the formula for molarity is

M= number of moles / volume    

6 = 0.019 / V

V=3.2 L

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