Name Section Pre Lab Questions 1. A certain bufferis made by disolving NaHICO, a

Question

Name Section Pre Lab Questions 1. A certain bufferis made by disolving NaHICO, and Na CO, in some wate. equation to show, how this bufer neutralizes added H+ and OH.. 2. Calculate the pH of 0.100M propanoic acid(HC, H,O, Ka-13X10) pH=-ba(o.ooli ) 2.. 96 3. Calculate the pH of 0.10 M HONH, (Kh=1.1x10-). ph , lf-foh-lut4.ng = 1-S 2 4. Calculate the pH after 0.020 mole HCl is added to 1.0 L of 0.100 M propanoic acid HC,H502, Ka=13X 105. 5. Calculate the pH of 0.1M acetic acid /0.25M Sodium acetate.

Explanation / Answer

1) The buffer contains NaHCO3 and Na2CO3. Sodium (Na+) is the spectator ion, hence, the actual buffer system comprises of HCO3- and CO32-. HCO3- is the acid component of the buffer (reacts with OH-) while CO32- is the basic component of the buffer (reacts with H+). The ionic equations for the reactions are

HCO3- (aq) + OH- (aq) -----------> CO32- (aq) + H2O (l)

CO32- (aq) + H+ (aq) ---------> HCO3- (aq)

Thus, the buffer can neutralize both added H+ and OH-.

2) Write down the acid ionization as

HC3H5O2 (aq) ---------> H+ (aq) + C3H5O2- (aq)

The ionization constant is given as

Ka = [H+][C3H5O2-]/[HC3H5O2] = (x).(x)/(0.100 – x)

Since Ka is small, we have (0.100 – x) 0.100; therefore,

Ka = x2/(0.100)

====> 1.3*10-5 = x2/(0.100)

====> x2 = 1.3*10-5*0.100 = 1.3*10-6

====> x = 1.140*10-3

Therefore, [H+] = 1.140*10-3 M and pH = -log [H+] = -log (1.140*10-3) = 2.9431 2.94 (ans).

3) HONH2 ionizes in water as below.

HONH2 (aq) + H2O (l) --------> HONH3+ (aq) + OH- (aq)

The base ionization constant is written as

Kb = [HONH3+][OH-]/[HONH2] = (x).(x)/(0.10 – x)

Since Kb is small, we have (0.10 – x) 0.10; therefore,

Kb = x2/(0.10)

=====> 1.1*10-8 = x2/(0.10)

=====> x2 = 1.1*10-8*0.10 = 1.1*10-9

=====> x = 3.317*10-5

Therefore, [OH-] = 3.317*10-5 M and pOH = -log [OH-] = -log (3.317*10-5) = 4.479.

We know that pH + pOH = 14; therefore, pH = 14 – pOH = 14 – 4.479 = 9.521 9.52 (ans).

4) Moles of propionic acid present = (1.0 L)*(0.100 M) = 0.100 mole.

We have already calculated the pH of 0.100 M propionic acid in (1). HCl is a strong acid and ionizes completely top produce H+. The volume of the solution remains unchanged; hence, the molarity of the added H+ = (0.020 mole)/(1 L) = 0.020 M.

Since HCl is a stronger acid, the pH is guided by HCl and is given as pH = -log [H+] = -log (0.020) = 1.699 1.70 (ans).

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