Student Worksheet for Buffers, Ksp, and Titrations

Transcription

1 Student Worksheet for Attempt to work the following practice problems after working through the sample problems in the videos. Answers are given on the last page(s). Relevant Equations Keq= [C]c [D] d [A] a [B] b ph= pk a + log [A ] [HA] [C] Ka = c [A] a [B] b pk a = -log Ka Ksp = Provided Value= [A] a [B] b ph= - log [H + ] = K eq [C][D] K a = 10 -pka s [A] or [B]= [aa] a =[bb] b M= moles Liter [H + ] = 10 -ph Quadratic Equation: ax 2 + bx + c * M1V1= M2V2 Where, A and B= reactants, C and D= products, a and b= coefficients of reactants, c and d= coefficients of products, s= solubility, Keq = equilibrium constant, Ksp= solubility constant, Ka = disassociation constant of the acid, A - = acid after donating its H, HA= acid before donating its H, M= molarity, and [ ] = concentration * The a, b, and c values for the quadratic equation come from the Ka equation Supercharged Science 1

2 1. If 100mL of 1.2M HCl are required to neutralize 75 ml of NaOH, what is the concentration of NaOH? 2. What is the ph of a solution that is a 3.5M weak, monoprotic acid that has a Ka of 1.2? 3. What is the solubility of Ag and CrO4, according to the following equation, if the Ksp of Ag2CrO4 is 1.1*10-12? 2 Ag + + CrO4 2- Ag2CrO4 4. Calculate the pk a of a weak acid solution that has a ph of 5.7 after the addition of 7 moles of the acid Supercharged Science 2

3 5. Calculate the solubility of Ag if the Ksp of AgBr is 5.4* Calculate the Ksp of Fe(OH)3 if the solubility of Fe 3+ and - OH are 1.29* What is the ph of a solution when 125 ml of a 4M acid solution has a pka of If 75mL of 2 M sulfuric acid is needed to neutralize Mg(OH)2, how many moles of Mg(OH)2 was in the original solution? Assume only 1 H + disassociates Supercharged Science 3

4 9. What is the ph of the solution in #8? 10. Benzoic Acid, C6H5COOH, is a weak acid with a Ka of 6.46*10-5. If 100 g are dissolved in 1.5L of water, what is the ph of the solution? NOTE: In this case, H2O acts as a base only. 11. The following chemical equation has a known Keq of What are the concentrations of products and reactants at equilibrium if 3 moles of CO are mixed with 5.7 moles of H2O? CO + H2O CO2 + H2 12. If the Ksp of Ca3(PO4)2 is 2.1*10-33, what is the concentration of Ca 2+ if it is dissolved in water to make a saturated solution? 2017 Supercharged Science 4

5 13. What is the Keq when HCl is mixed with S2 according to the following equation? 4 HCl + S2 2 H2S + 2 Cl2 14. Explain the observation that phosphate buffers have multiple equivalence points according to the following deprotonations: H3PO4 H2PO4 - + H + H2PO4 HPO H + HPO4 2- PO H Calculate the Ka of an acid if dissolving 1.5L of the 3M acid in 3 liters of water produces a solution that has a ph of Supercharged Science 5

6 1. If 100mL of 1.2M HCl are required to neutralize 75 ml of NaOH, what is the concentration of NaOH? You will first need to set up your balanced equation. This becomes: HCl + NaOH H2O + NaCl Now use the molarity formula to find out how many moles of HCl were actually added. M = moles Liter Moles = M * Liters You are solving for moles. Don t forget to convert ml to L. = 1.2 * 0.1 = 0.12 moles HCl added. From the equation, we know that there s a 1:1 ratio, so there were also 0.12 moles of NaOH. Plugging this into the molarity formula: M = moles Liter = = 1.6M of NaOH 2. What is the ph of a solution that is a 3.5M weak, monoprotic acid that has a Ka of 1.2? Because it is a weak acid, you will need to set up your ICE table. Ka = [B][C] [A] HA H + A - I C - X +X +X E 3.5-X +X +X Note: In this example, there is one reactant (the acid) and 2 products. 1.2 = X X 3.5 X = X X Set all variables equal to 0. X X 4.2= 0 Use the Quadratic Equation to solve for X. X= 1.54, The negative number can never be the answer because it implies that matter is destroyed. Now plug your values of X (1.54) into the H + space of the Equilibrium row in the table. [H + ] = 1.54 Use ph= -log [H + ] to solve ph= (Rare, but yes, that s possible) 2017 Supercharged Science 6

7 3. What is the solubility of Ag and CrO4, according to the following equation, if the Ksp of Ag2CrO4 is 1.1*10-12? 2 Ag + + CrO4 2- Ag2CrO4 Ksp = [Ag] 2 [CrO4 2- ]= 1.1*10-12 s 3 = 1.1*10-12 s= 1.03*10-4 each Note: Because there is twice as much Ag, you will multiply the s value by 2 to get the total solubility of Ag. This value is per ion. 4. Calculate the pk a of a weak acid solution that has a ph of 5.7 after the addition of 7 moles of the acid. Using the Henderson-Hasselbalch equation, ph = pk a + log [A ], you can use the reverse log [HA] formula [H + ]= 10 -ph. Because the A - value is equal to the H + value, these can plug straight in. [H + ]= = 2.0*10-6. Plugging the values into the Henderson equation, we get: ph = pka + log [A ] [HA] 5.7 = pka + log ( ) 5.7 log = pka pka = Calculate the solubility of Ag if the Ksp of AgBr is 5.4* Ksp = [Ag][Br]= 5.4*10-13 You probably guessed that it s the square root of the Ksp. s 2 = 5.4*10-13 = = 7.35* Supercharged Science 7

8 6. Calculate the Ksp of Fe(OH)3 if the solubility of Fe 3+ and - OH are 1.29* Ksp= s[a] a *s[b] b = 1.29*10-39 * (1.29*10-39 ) 3 = 2.8*10-39 mol/l 7. What is the ph of a solution when 125 ml of a 4M acid solution that has a pka of 1.9 is diluted to 1 L. Use the Molarity equation to find the moles of acid. M= moles Liter Moles= Molarity * Liter = 0.5 moles in 1 liter ph= -log (0.5) = If 75mL of 2 M sulfuric acid is needed to neutralize Mg(OH)2, how many moles of Mg(OH)2 was in the original solution? Assume only 1 H + disassociates. Write the balanced equation, and then use the molarity equation to determine the number of moles. M= moles Liter Moles= Molarity * Liter = 2* 0.75 = 1.5 moles H2SO4 2 H2SO4 + Mg(OH)2 2 HSO H2O + Mg 2+ From the equation, we know that there is a 2:1 molar ratio between the reactants. So, the number of moles of Mg(OH)2 is = 0.75 moles of Mg(OH) Supercharged Science 8

9 9. What is the ph of the solution in #8? Because all H + and - OH are neutralized, the ph is Benzoic Acid, C6H5COOH, is a weak acid with a Ka of 6.46*10-5. If 100 g are dissolved in 1.5L of water, what is the ph of the solution? NOTE: In this case, H2O acts as a base only. First find the moles of benzoic acid. Moles of Benzoic Acid= 100 = 0.82 moles 122 C6H5COOH + H2O H3O + + C6H5COO - C6H5COOH H3O + C6H5COO - I C -X +X +X E 0.82-X X X 6.46*10-5 = [H 3O + ][C 6H5 COO ] [C 6H5 COOH] = X X Multiply both sides by the denominator and distribute 5.3* *10-5 X = X 2 Set everything equal to 0. X * *10-5 = 0 Solve for X via the Quadratic Equation X= 7.36*10-6, -7.2*10-5 The H + concentration is 7.36*10-6. ph= -log H + = 5.13 Note: In this example, the moles of water are not included in the ICE table because there is significantly more of it. Therefore, it s change is considered to be negligible. 11. The following chemical equation has a known Keq of What are the concentrations of products and reactants at equilibrium if 3 moles of CO are mixed with 5.7 moles of H2O? CO + H2O CO2 + H2 CO H2O CO2 H2 I C -X -X +X +X E 3-X 5.7-X X X 0.64 = [CO 2 ][H 2] [CO][H 2 O] = X X (3 X)(5.7 X) Multiply both sides, FOIL, and distribute Supercharged Science 9

10 0.64X 2 = X 5.7X + X 2 Combine like terms. 0.64X 2 5.6X = X 2 Set everything equal to X X 10.9 = 0 Solve for X via the Quadratic Equation. X = 1.75, [CO] = = 1.25 moles [H2O] = = 3.95 moles [CO2] = 1.75 moles [H2] = 1.75 moles 12. If the Ksp of Ca3(PO4)2 is 2.1*10-33, what is the concentration of Ca 2+ if it is dissolved in water to make a saturated solution? Start by writing the disassociation equation. Ca3(PO4)2 3 Ca PO *10-33 = (3s) 3 (2s) 2 = 27s 3 (4s 2 ) = 108s 5 5 = [Ca 2+ ] = 1.14*10-7 moles/l 13. What is the Keq when HCl is mixed with S2 according to the following equation? Keq = [C]c [D] d [A] a [B] b = = 1 mol/l 4 HCl + S2 2 H2S + 2 Cl2 14. Explain the observation that phosphate buffers have multiple equivalence points according to the following deprotonations: H3PO4 H2PO4 - + H + H2PO4 HPO H + HPO4 2- PO H Supercharged Science 10

11 The equivalence point is when the concentration of H + equals the concentration of each phosphate compound (left side of the reaction arrow above). H3PO4 has 3 hydrogens attached, which lends to three different areas of the titration graph that H + will equal phosphates. 15. Calculate the Ka of an acid if dissolving 1.5L of the 3M acid in 3 liters of water produces a solution that has a ph of [H + ] = 10 -ph = = 1.91*10-6 (This value is equal to the base after deprotonation) Start by using the Molarity dilution formula to calculate the final molarity of solution. M1V1 = M2V2 M2 = M 1V 1 V 2 = 1.5L3M 4.5L = 1M Now calculate the number of moles present. 1M = X moles 4.5L X moles = 4.5L * 1 M = 4.5 moles Now use the Henderson-Hasselbalch Equation to solve for pka: ph= pka + log [A ] [HA] pka = ph - log [A ] [HA] = log [ ] [4.5] = (-6.37) = Ka = 10 -pka = 8.13* Supercharged Science 11