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Transcript
Autoionization of Water Although water is a molecular substance, very low concentrations of hydronium ions and hydroxides ions are formed by autoionization. H O H + H O H H O - + H O+ H H Autoinization Autoionization of Water Autoionization: the process in which water spontaneously forms low concentrations of H+ and OH - ions by proton transfer from one water molecule to another 2 H2O (l) H3O+ (aq) + OH- (aq) At any given time only a very small number of water molecules are ionized. • If every letter in our text represented a water molecule, you would have to look through about 50 texts to find one H3O+! Autoionization of Water An equilibrium constant expression can be written for the autoionization of water: 2 H2O (l) H3O+ (aq) + OH - (aq) Kw = [H3O+] [OH -] = [H+ ] [OH- ] where Kw = ionization constant for water = ion-product constant = 1.00 x 10-14 at 25oC Autoionization of Water The value of Kw (and all other equilibrium constants) varies with temperature: Kw = 1.14 x 10-15 at 0oC Kw = 1.00 x 10-14 at 25oC Kw = 9.61 x 10-14 at 60oC You should assume 25oC unless otherwise stated in a problem. Autoionization of Water By definition: Neutral solution: [H+ ] = [OH- ] Acidic solution: [H+ ] > [OH- ] Basic solution: [H+ ] < [OH- ] Autoionization of Water Example: Calculate the [H+ ] and [OH- ] in a neutral solution at 25oC. Autoionization of Water Example: What is the [H+ ] at 25oC for a solution in which [OH- ] = 0.010 M. Autoionization of Water Example: What is the [OH- ] at 25oC in a solution in which [H+ ] = 2.5 x 10-6 M. pH Since the [H+] is usually very small in aqueous solutions, we normally express the [H+] in terms of pH. pH = - log10 [H+] Sig Figs and logs: only the digits after the decimal point are significant. pH Example: Calculate the pH of a solution with [H+] = 2.52 x 10-5. pH Example: Calculate the pH of a solution with [OH-] = 6.5 x 10-5. pH The negative log is also used to express the magnitude of other small quantities: pOH = - log [OH- ] pH and pOH are related by the following equation that is derived by taking the negative log of the expression for Kw pH + pOH = 14.00 at 25oC pH Example: Calculate the pOH of a solution with [OH - ] = 2.5 x 10-3 M. pH Example: Calculate the pH of a solution with [OH - ] = 2.5 x 10-3 M. Approach 1: pH Approach # 2: pH Given the pH of a solution, you can also find the [H+] and the [OH-]. Since pH = - log [H+], [H+] = 10 –pH Since pOH = - log [OH-], [OH-] = 10 -pOH pH Example: What are the [H+] and [OH-] for a solution with a pH of 2.50 at 25oC? Strong Acids Strong acid: Strong electrolyte Ionizes completely in aqueous solution HNO3 (aq) H + (aq) + NO3 – (aq) The only significant source of H+ ion in an aqueous solution of a strong acid is usually the strong acid. Strong Acids Consequently, the [H+] in a solution of a strong monoprotic acid can be determined easily using the concentration of the strong acid itself. HNO3 (aq) H + (aq) + NO3 – (aq) In a 0.05 M HNO3 (aq) solution, [H+] = 0.05 mol HNO3 x 1 mol H+ L 1 mol HNO3 = 0.05 M Strong Acids Example: What is the pH of a 0.25 M HCl (aq) solution? Strong Bases Strong Base strong electrolyte ionizes completely in aqueous solution NaOH (aq) Na+ (aq) + OH- (aq) Common strong bases alkali metal hydroxides hydroxides of Ca, Sr, and Ba Strong Bases The pH of an aqueous solution of a strong base can be determined using the concentration of the strong base NaOH (aq) Na+ (aq) + OH- (aq) A 0.25 M solution of NaOH has an [OH-] of 0.25 M: 0.25 mol NaOH x 1 mol OHL 1 mol NaOH = 0.25 M Strong Bases The pH of the base solution can then be found in two ways: Calculate pOH use pH + pOH = 14.00 to determine pH Calculate [H+] use [H+] [OH-] = 1.00 x 10-14 Then calculate pH Strong Bases Example: Calculate the pH of a 0.25 M Ca(OH)2 (aq) solution. Step 1: Determine [OH-] Strong Bases Step 2: Calculate pOH Step 3: Calculate pH Strong Bases Example: What is the pH of a solution prepared by mixing 10.0 mL of 0.015 M Ba(OH)2 and 30.0 mL of 7.5 x 10-3 M NaOH? Strong Bases Step 1: Find the total [OH-] Strong Bases Step 2: Find pOH Step 3: Find pH Weak Acids Most acidic substances are weak acids: partially ionize in solution the solution contains an equilibrium mixture of acid molecules and its component ions CH3CO2H H+ (aq) + CH3CO2 - (aq) Weak Acids The extent to which a weak acid ionizes can be expressed using an equilibrium constant known as the acid-dissociation constant (Ka). For a general reaction: HX (aq) H+ (aq) + X- (aq) Ka = [H+][X-] [HX] Note: The rules for writing an expression for Ka are the same as those for Kc, Kp and Ksp. Weak Acids The magnitude of Ka indicates the tendency of the hydrogen ion in an acid to ionize. The larger the value of Ka, the stronger the acid is. The pH of a weak acid solution can be calculated using the initial concentration of the weak acid and its Ka. Weak Acids To calculate the pH of a weak acid solution: Write the ionization equilibrium for the acid. Write the equilibrium constant expression and its numerical value. Set up a table showing initial concentration, change, equilibrium concentration. Substitute equilibrium concentrations into the equilibrium constant expression. Weak Acids To calculate the pH of a weak acid solution (cont): Solve for the change in concentration. Assume that the change in concentration is small (i.e. < 5%) compared to the initial concentration of the weak acid. Check the validity of previous assumptions. If x/initial concentration x 100% >5.0%, you must use the quadratic equation to solve for x. Calculate the final concentrations and pH. Weak Acids Example: Calculate the pH of a 0.20 M solution of HCN. Ka = 4.9 x 10-10 Step 1: Write the equation for the ionization. Step 2: Write the expression for Ka. Weak Acids Step 3: Set up a table. Weak Acids Step 4: Substitute equilibrium concentrations into the Ka expression. Step 5: Assume that x << 0.20 M and solve for x. Weak Acids Step 6: Check the validity of our assumption. Weak Acids Step 7: Substitute value for x into the table to find the [H+]. Weak Acids Step 8: Calculate the pH using the [H+] Weak Bases Many substances behave as weak bases in water. Weak base + H2O conjugate acid + OH- The extent to which a weak base reacts with water to form its conjugate acid and OH- ion can be expressed using an equilibrium constant known as the base-dissociation constant (Kb). Weak Bases Kb always refers to the equilibrium in which a base reacts with water to form its conjugate acid and OH- ion. For the reaction: NH3 (aq) + H2O (l) Kb = [NH4+] [OH-] [NH3] NH4+ (aq) + OH- (aq) Note: The rules for writing an expression for Kb are the same as those for Kc, Kp and Ksp. Weak Base To calculate the pH of a weak base solution: Write the ionization equilibrium for the base. Write the equilibrium constant expression and its numerical value. Set up a table showing initial concentration, change, equilibrium concentration. Substitute equilibrium concentrations into the equilibrium constant expression. Weak Bases To calculate the pH of a weak base solution (cont): Solve for the change in concentration. Assume that the change in concentration is small (i.e. < 5%) compared to the initial concentration of the weak base. Check the validity of previous assumption. Calculate the [OH-] concentration and pOH Use pOH to calculate pH. Weak Bases Example: Calculate the pH of a 0.20 M solution of methylamine, CH3NH2. Kb = 3.6 x 10-4. Step 1: Write the equation for the ionization. Step 2: Write the expression for Kb. Weak Bases Step 3: Set up a table. Weak Bases Step 4: Substitute equilibrium concentrations into the Kb expression. Step 5: Assume that x << 0.20 M and solve for x. Weak Bases Step 6: Check the validity of our assumption. Weak Bases Step 7: Substitute value for x into the table to find the [OH-]. Weak Acids Step 8: Calculate the pOH Step 9: Calculate the pH