Question

calculate the number of grams of CH3COONa * 3H2O (sodium acetate tri-hydrate) needed to make 250.0 mL of a CH3COOH (acetic acid)/ CH3COONa * 3H2O buffer. The target pH of the buffer is 5.25. The given concentration of [CH3COOH] is equal to 0.10 M. Ka = 1.80 x 10-5 for acetic acid.

Use the Henderson-Hasselbalch equation to determine the ratio of [CH3COO-]/[CH3COOH] needed to make the buffer. pH = pKa + log([CH3COO-]/[CH3COOH])

Now calculate the concentration of acetate ion (CH3COO- ) needed to make the buffer. Hint: Recall the concentration of CH3COOH given in the instructions.

Now, give the concentration of CH3COONa * 3H2O needed to make the buffer. Hint: The CH3COONa * 3H2O dissolves in water to form Na+ ions and CH3COO-.

Finally, convert the concentration of CH3COONa * 3H2O into grams of CH3COONa * 3H2O needed to make the buffer.

Answer 1

solution - for acidic Buffer pH = pka + log [salt] pka_-logka "Cacid] 1.18 = -log (t. so x 10) 5-25 = 4.75 + dog csalt) * o. Lo Eldorsens į 4.75 time = * log [salt 0.5 color= 6 - By taking antilogi.com and Olio (salt) con 3 com = !! 4 . [orto faritazhdo] [salt] ñ 0.30 [salt),= L. 60 + salt (molar mass) ( solution ) 0:30 = (cat)salt x 1080 4 (136) C250 m). 10.30 136 - 10.2 (wt Salt - gram Scanned with CamScanner 4