Determine the hydroxide ion concentration from the K b and then determine the

# Determine the hydroxide ion concentration from the k

• Notes
• 64

This preview shows page 28 - 30 out of 64 pages.

Determine the hydroxide ion concentration from the K b , and then determine the pH from the pOH. For the first equivalence point: K b = 2.222x10 –8 = 2 3 3 H CO OH HCO =    x x 0.0404875 x =    x x 0.0404875 [OH ] = x = 2.999387x10 –5 M pOH = – log (2.999387x10 –5 ) = 4.522967495 pH = 14.00 – pOH = 14.00 – 4.522967495 = 9.4770 = 9.48 For the second equivalence point: K b = 2.1276596x10 –4 = 3 2 3 HCO OH CO     =    x x 0.023977 x =  x x 0.023977 [OH ] = x = 2.2586477x10 –3 M pOH = – log (2.2586477x10 –3 ) = 2.6461515 pH = 14.00 – pOH = 14.00 – 2.6461515 = 11.3538 = 11.35 19.60 Plan: Use ( M )( V ) to find the initial moles of base and then use the mole ratio in the balanced equation to find moles of acid; dividing moles of acid by the molarity of the acid gives the volume. At the equivalence point, the conjugate acid of the weak base is present; set up a reaction table for the acid dissociation in which x = the amount of dissociated acid. Use the K a expression to solve for x from which pH is obtained. Solution: a) The balanced chemical equation is: HCl( aq ) + NH 3 ( aq ) NH 4 + ( aq ) + Cl ( aq ) The chloride ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of HCl needed: Volume (mL) of HCl = 3 3 3 3 0.234 mol NH 10 L 1 mol HCl L 1 mL 65.5 mL L 1 mL 1 mol NH 0.125 mol HCl 10 L   = 122.616 = 123 mL HCl Determine the moles of NH 3 present: Moles = 3 3 0.234 mol NH 10 L 65.5 mL L 1 mL = 0.015327 mol NH 3 At the equivalence point, 0.015327 mol HCl will be added so the moles acid = moles base. The HCl will react with an equal amount of the base, 0 mol NH 3 will remain, and 0.015327 moles of NH 4 + will be formed. HCl( aq ) + NH 3 ( aq ) NH 4 + ( aq ) + Cl ( aq ) Initial 0.015327 mol 0.015327 mol 0 Change –0.015327 mol –0.015327 mol +0.015327 mol Final 0 0 0.015327 mol Determine the liters of solution present at the equivalence point: Volume = [(65.5 + 122.616) mL](10 –3 L/1 mL) = 0.188116 L Concentration of NH 4 + at equivalence point:
19-29 Molarity = (0.015327 mol NH 4 + )/(0.188116 L) = 0.081476 M Calculate K a for NH 4 + : K b NH 3 = 1.76x10 –5 K a = K w / K b = (1.0x10 –14 )/(1.76x10 –5 ) = 5.6818x10 –10 Using a reaction table for the equilibrium reaction of NH 4 + : NH 4 + + H 2 O NH 3 + H 3 O + Initial 0.081476 M 0 0 Change –x +x +x Equilibrium 0.081476 – x x x Determine the hydrogen ion concentration from the K a , and then determine the pH. K a = 5.6818x10 –10 = 3 3 4 H O NH NH =    x x 0.081476 x =  x x 0.081476 x = [H 3 O + ] = 6.803898x10 –6 M pH = –log [H 3 O + ] = –log (6.803898x10 –6 ) = 5.1672 = 5.17 b) The balanced chemical equation is: HCl( aq ) + CH 3 NH 2 ( aq ) CH 3 NH 3 + ( aq ) + Cl ( aq ) The chloride ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of HCl needed: Volume (mL) of HCl = 3 3 2 3 3 2 1.11 mol CH NH 10 L 1 mol HCl L 1 mL 21.8 mL L 1 mL 1 mol CH NH 0.125 mol HCl 10 L