Question

1) What is the pH when 5.00 mL of 0.200 M NaOH has been added to 50.00 mL of 0.100 M CH3CO2?

Strategy

Step 1: Use stoichiometry of the neutralization to determine the amounts of acid and conjugate base present in solution

Step 2: Solve for equilibrium concentrations using ICE tables or Henderson-Hasselbalch approximation

2) Calculate the molar solubility of silver chromate in pure water. K_sp = 1.12 x 10¯¹²

3) Calculate K_b for the acetate ion from the K_a for acetic acid of 1.80 x 10^-5.

4) Sketch four different titration curves including at least one for a polyprotic acid. On each curve indicate:

1. pH on one axis and concentration of _____________ on other axis.
2. Strength of acid and base
3. Equivalence point
4. Buffer region (if relevant)

Answer 1

1 Answer is

- we know that number of moles (n) = Molarity x Volleme (in my 10oo sn= MV Cinmi) L 1000 NaoH moks: V.Sml M =0.2M ņ= (0:2) 100o = 1x10- Сн сос, тоо | V = Somì, 1 = 0.1M na 50 (0:1) 1000 - S X The stoichiometry of neutralization reaction is ch cool + NaOH = CH CooNa + ho Acid Conjugate base I mole CHCOOH Produce 1 vole conjugate base CCH Coona). cs Scanned with lewis CamScanner

ICE table Ch Cook + NaOH - Initially: 5x10-3 1x103 change - 1x10-3 -1X1o-3 Finally: 4x10- 3 0 Ch cooNat the 0 0 Ixlos - From Henderson - Hassel balch equation P" = pka + log number of moles of salt l number of moles of acid number of moles of salt = number of moles of conjugate base = 1x103 number of moles of acid = 4X103 pka of acetic acid=4.95 substitute the values, then pH = 4.75+ log (long) pH = 4.75 +log (14) pH = 4.75 -0.602 pH = 4.15 1 s Sinned with and camscanner