How many moles of sodium acetate (nach3coo) must be added to 1.000 liter of a 0.500 m solution of acetic acid (ch3cooh) to produce a ph of 5.061? the ionization constant of acetic acid is 1.8 ã 10â5 . answer in units of mol?

When PH = -㏒[H3O+] 
and we have PH = 5.061 
by substitution:
∴ [ H3O+] = 10^-5.061 = 8.7x10^-6
when we have Ka = [CH3COO-][H3O+] / [CH3COOH]
we have Ka = 1.8x10^-5 & [H3O+] = 8.7x10^-6m & [CH3COOH] = 0.5 m 
by substitution in Ka formula:
1.8x10^-5 = [CH3COO-]*(8.7x10^-6) / 0.5
∴[CH3COO-] = 1.034
∴we need  to add 1.034 mol of sodium acetate

The answer would be true

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