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Problem 1 - What is the pH of each of the following solutions a. 0.04 M HCl Strong acids completely dissociate in so

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Problem 1 - What is the pH of each of the following solutions a. 0.04 M HCl Strong acids completely dissociate in so

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    1. Problem #1 - What is the pH of each of the following solutions? a). 0.04 M HCl Strong acids completely dissociate in solution therefore the total concentration is equal to the concentration of H+. pH = -log[H+] = -log [0.04 M] = 1.4

    2. Problem #1 - What is the pH of each of the following solutions? b). 0.04 M NaH2PO4 The dissociation of weak acids in solution is not complete so you must first calculate the concentration of the H+ using the Ka for the weak acid. Ka for NaH2PO4= 6.23 x 10-8 M [H+]2 [H+]2 Ka = ––––– = –––– = 6.23 x 10-8 M [HA] [0.04] [H+] = v(Ka)(Total concentration) = v(6.23 x 10-8 M)(0.04) = 5.0 x 10-5 M pH = -log [5 x 10-5 M] = 4.3

    3. Problem #1 - What is the pH of each of the following solutions? c). 0.1 M H2CO3 (HA) and 0.6 M NaHCO3 (A-) pH = pKa + log [A-] [HA] pH = 6.4 + log (0.6/0.1) = 7.2 (H2CO3 pKa=6.4)

    4. Problem #1 - What is the pH of each of the following solutions? d). 0.1 M H2CO3 (HA) and 0.08 M NaOH (acceptor) pH = pKa + log [A-] [HA] pH = 6.4 + log 0.08/(0.1 – 0.08) = 7.0 (NaOH will completely dissociate) (H2CO3 pKa=6.4) Here you are figuring out to find out how much of the buffer was used up by the base (NaOH).

    5. Problem #1 - What is the pH of each of the following solutions? e). 0.45 M NaHCO3 (A-)and 0.35 M HCl (donor) pH = pKa + log [A-]/[HA] = 6.4 + log (0.45 – 0.35)/0.35 = 5.9 (HCl will completely dissociate) One of the forms (not balanced): NaHCO3 ->H2CO3 So use pKa of H2CO3 (6.4) HA + OH- -> A + H2O A- + H+ -> HA Here you are figuring out to find out how much of the buffer was used up by the acid (HCl).

    6. Problem #2 - Determine the H+ and OH- concentration in cow’s milk. The pH of cow’s milk is typically 6.5. The [H+] can be determined from the equation pH = -log [H+]. 6.5 = - log [H+] or 10-6.5 = [H+] = 3.2 x 10-7M Since the ion product of water is Kw= [H+][OH-] 1 x 10-14=[H+][OH-] [OH-] = 1 x 10-14 /3.2 x 10-7 = 3.2 x 10-8M.

    7. Problem #3 - Sketch a titration curve for tyrosine. What is the isoelectric point of tyrosine? Use table 3.2 for pKa values.

    9. Problem #4 – How many moles of NaOH would you need to add to 800 ml of a 0.4 M H2CO3 solution in order to bring the pH of the solution to 10.5? We have 0.32 moles of H2CO3 (800 ml of 0.4 M) and we need to figure out how much NaOH needs to be added to get a final pH of 10.5 When you add 0.32 moles of (equal amount) you convert all the H2CO3 to HCO3- Next, find the ratio of HCO3- [HA] and CO3-- [A-] needed to achieve a pH of 10.5: pKa HCO3- = 10.2 10.5 = 10.2 + log[A-]/[HA] 0.3 = log[A-]/[HA] 10^(0.3) = 10^(log[A-]/[HA]) ? 2 = [A-]/[HA] 0.32 = [HA] + [A-] or [A-] = 0.32 - [HA] 2 = [A-]/[HA] So 2 = (0.32 - [HA])/[HA] ? [HA] = 0.103 0.32 = 0.103 + [A-] [A-] = 0.32 - 0.103 [A-] = 0.21 moles We need to add another 0.21 moles of NaOH to our starting amount (0.32 mol) So 0.21 moles + 0.32 moles = 0.53 moles will yield a pH of 10.5.

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