Lecture 8, Chemistry of Solutions Notes
Lecture 8, Chemistry of Solutions Notes 202-NYB-05
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This 5 page Class Notes was uploaded by CatLover44 on Saturday October 8, 2016. The Class Notes belongs to 202-NYB-05 at Dawson Community College taught by Nadia Schoonhoven in Fall 2016. Since its upload, it has received 7 views. For similar materials see Chemistry of Solutions in Chemistry at Dawson Community College.
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Date Created: 10/08/16
Lecture 8: Tuesday, September 20, 2016 Chemistry of Solutions Prof.: N. Schoonhoven Topics covered: equilibrium constants for acids and bases, amphoteric substances, auto-ionization of water, the pH scale, the pH of strong acids. Equilibrium constants for acids and bases For eqm involving an acid reacting with water, K . a Acids with K > 1 are strong, and acids with K < 1 are weak. Water isn't written in eqm a a constants for acids or bases. - + K a [A][H O ]3/ [HA] So a strong acid dissociates according to the following general equation: HA + H O ⇌ A - + H O + or HA ⇌ H + + A - 2 (l) (aq) 3 (aq) (aq) (aq) K as of strong acids are too large to measure in H O 2ecause they dissociate almost completely. Amphoteric substances Some substances are amphoteric, meaning they can behave as an acid and a base. Water is an amphoteric substance. + - The K oa water is referred to as K , whiwh is written as K = [H ][Ow ], and has a value of -14 + - -14 1.0x10 at 25.0°C. In aqueous solutions at 25.0°C, [H ] x [OH ] must equal 1.0x10 . Example: for a solution at 25.0°C, [H ] = 1.0x10 M. What is [OH ]? - K w [H ][OH ] - (1.0x10 ) / (1.0x10 ) = 1.0x10 M = [OH ]. - Auto-Ionisation of Water Energy + 2H O ⇌2H (l) 3 +(aq)+ OH -(aq)(endothermic). -13 -14 K wt 60.0°C = 6.0x10 , K at 25.0wC is 1.0x10 . Example: In the dissociation equation for aniline, C H NH , 2ha5 is t2e conjugate acid? C 2 N5 ⇌ C2H NH 2 O5 3+ - The conjugate acid is C H NH . + 2 5 3 Nitric acid, HNO , i3 considered to be a strong acid whereas nitrous acid, HNO , is considere2 to - be a weak acid. Nitric acid has an aqueous eqm position that lies far to the right, and NO is 3 considered a weak conjugate base. Example: use Table 14.2 to determine which of the following bases would have the weakest conjugate acid. 2 Write the dissociation equations for each base: - + - 1. C H 2 +3H 2 ⇌ HC 2 O + OH 2 3 2 2. NH + H3O ⇌ N24 + OH + - 3. ClO + H O ⇌ 2ClO + OH - 4. SO 42-+ H 2 ⇌ H SO 2 OH4 - + NH4 has the smallest K , so itas the weakest conjugate acid. + -6 - Example: in a solution of water at 25°C, [H ] = 1.2x10 M. What's the [OH ] in the solution, and is it an acidic or basic solution? -14 -6 -8 -9 (1.0x10 ) / (1.2x10 ) = (0.83)x10 = 8.3x10 M - -9 -7 [OH ] = 8.3x10 M < 1.0x10 M so the solution is acidic. The pH Scale If [H ] = 1.0x10 M, pH = 10 = 7.00 (pH has to have as many decimal spaces as there are sig figs in the concentration value). pOH = -log[OH ], and pK = -logK. In pure water, [H ] = [OH ] = 1.0x10 M, and pH + pOH = 14. Example: the pH of a soft drink is approximately 3.30, so: + -3.30 -4 [H ] = 1.0x10 = 5.0x10 M. - -10.7 -11 And pH + pOH = 14, so pOH = 14 – 3.3 = 10.7, which means [OH ] = 10 = 2.0x10 M. -3 + Example: you have a 1.0x10 M solution of NaOH at 25°C. What's the [H ] in this solution? (Without using a calculator). -3 pOH = 10 = 3 pH = 14 – 3 = 11 So [H ] = 1.0x10 -11M. 3 pH and pOH K w [H ][OH ] - Consider the log form of K : w log K w log [H ] + log [OH ] - -log K w -log [H ] – log [OH ] - pH = pH + pOH, since K is 1w0x10 -14at 25°C. - + -6 Example: determine the pH, pOH, and [OH ] of an aqueous solution with [H ] = 3.5x10 M. -6 -log(3.5x10 ) = 5.5 = pH pH + pOH = 14 14 – 5.5 = 8.5 -8.5 -9 - 10 = 3.0x10 M = [OH ] Note: a weak acid with a greater concentration than that of a strong acid could still have a lower pH than the strong acid. pH of Strong Acid Solutions - Always consider the species present in a solution. For example, a solution of HCl contains + - mostly H (aq)and Cl (aq) Example: what is the pH of 1.0 M HCl? Determine the sources of H +(aq) HCl is a strong acid, so ~100% ionization. 4 + - HCl → H + Cl . pH = -log(1.0) = 0. -10 Example: calculate the pH of a 1.0x10 M HNO sol3tion. + - HNO → 3 + NO 3 HNO is3a strong acid, so [HNO ] ⇌ [H 3 = 1.0x10 -10M. + - + -7 -10 H 2 ⇌ H + OH , [H ] = 1.0x10 M > 1.0x10 M. + -7 So [H ] = 1.0x10 M in the HNO solution3 Note: in strong acids, the contribution of H+ from water is usually insignificant, the concentration of the acid is very low, and strong acids dissociate almost completely (assume 100% dissociation). Example: A 0.040 M solution of an acid, HA, has a pH of 3.02 at 25°C. What is the Ka for this acid? [H ] = 10 -3.0= 9.5x10 M = [A]. - Ka = (9.5x10 ) / (4.0x10 ) -2 -5 Ka = 2.3x10 5
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