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. PLEASE EXPLAIN :(

Q1:Given the K_a value in the instructions (1.80 x 10^-5). Calculate the pK_a

Q2:Use the Henderson-Hasselbalch equation to determine the ratio of [CH3COO-]/[CH3COOH] needed to make the buffer. pH = pKa + log([CH3COO-]/[CH3COOH])

Q3:Now calculate the concentration of acetate ion (CH3COO- ) needed to make the buffer. Hint: Recall the concentration of CH3COOH given in the instructions.

Q4: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-.

Q5:Finally, convert the concentration of CH3COONa * 3H2O into grams of CH3COONa * 3H2O needed to make the buffer.

Answer 1

Acidic buffers are solutions that have a pH below 7 and contain a weak acid and one of its salts. For example, a mixture of acetic acid and sodium acetate acts as a buffer solution
ph for acidic burger is given by - Pro plat 10). Content Acetic Acida 1.8x105 az-logna ka = 4.76 pH=5.25 >> 5.25-4.763 log [salt] [acid] Cid => 3:09 [sace [acid

Ten, coon] - [acid = o.lom [salt] = 0.30am No of moles of salt - volume of solution in 0.309 m . wo or moles xlovo - 0309m 250 No or moles of faitz 0.309 20.077 No of moles of saltz No of moles of CH LOONa. 34₂0 20.077 Mw or ch₂ COONa. 340 = 82 +54 = 1369 So, mass or chCOONa. no needed=1368 0.0779 = 10.48297