how to prepare a solution?

Question

of ph 4.5 using nach3coo in 800 ml water find no og grams of nach3coo needed ka is 1.8*10^-5

Answer 1

CH3COONa is the salt of a weak acid (CH3COOH) and a strong base. Thus it will hydrolyze giving an alkaline solution.
In order to have acidic pH you will need to add acid. Thus some CH3COOH will form and you will have the buffer CH3COOH-CH3COONa. Using the Henderson-Hasselbalch equation

pH=pKa + log([CH3COONa] / [CH3COOH])=>
4.5 =-log(1.8*10^-5) + log([CH3COONa] / [CH3COOH])=>
4.5 =4.7 + log([CH3COONa] / [CH3COOH])=>
-0.2= + log([CH3COONa] / [CH3COOH])=>
[CH3COONa] / [CH3COOH] = 0.631

Since we don't know the final concentration of the buffer (C=[CH3COONa] + [CH3COOH]) or have any other info it is not possible to continue and find the grams of CH3COONa needed.