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230_Sp10_E3_key(1) - 1(20 pts 20.0 mL of a 0.1M aqueous...

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1) ( 20 pts ) 20.0 mL of a 0.1M aqueous acetic acid (pKa = 4.75) solution is titrated with 0.1 M NaOH. Find the exact values of the coordinates for the three points listed in the table below: 1) half-stoichiometric point; 2) the stoichiometric point; 3) the point where 23mL of the NaOH have been added. Place those points on the graph and sketch the correct shape of the curve through them. 1 pts each for volume and 2pts pH=4.75, 6 pts for pH=8.72, 12.8; 1pt for points on graph; 3 points for shape of curve mL NaOH added pH 1) ½ stoichiometric point 10 4.75 2) stoichiometric point 20 8.72 3) 23 11.84 Show your work below: 0.02 L ( ) 0.1 molHA L " # $ % & = 0.002 molHA At stoich point molHA = mol OH - 0.002 molesOH ( = xL 0.1 molOH ( L " # $ % & 0.02 L = x = 20 mL (1/2 stoich = 10 mL ; pH = pKa = 4.75 at stoichiometric point : HA + OH - A - + H 2 O P 0.002 0.002 0 0 U-0.002 -0.002 + 0.002 G 0 0 0.002mol/0.04L A - + H 2 O HA + OH - I 0.05M 0 0 C -x +x +x E 0.05-x x x pK a = 4.75 K a = 10 " 4.75 K b = 1 x 10 " 14 10 " 4.75 = 5.62 x 10 " 10 = x 2 0.05 " x shortcut , x = 5.3 x 10 " 6 = 0.01% of 0.05 = [ OH " ] pOH = " log(5.3 x 10 " 6 ) pH = 14 " pOH = 8.72 @23mL NaOH added 0.023 L ( ) 0.1 molOH " L # $ % & ( = 0.0023 mol HA + OH - A - + H 2 O P 0.002 0.0023 0 0 U-0.002 -0.002 + 0.002 G 0 0.0003 0.002mol [OH - ]=0.0003mol/0.043L=6.98x10 -3 M pOH=-log[6.98x10 -2 M]=2.16 pH = 11.84 Fill in the columns for mL NaOH added and pH in the table. On graph plot 3 points and draw shape of the titration curve.
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3) An aqueous solution of Sr(NO 3 ) 2 is added slowly to 1.0 L of a well-stirred solution containing 0.020 mole F - and 0.10 mole SO 4 2- at 25ºC. (You may assume that the added Sr(NO 3 ) 2 solution does not affect the total volume of the system.) SrSO 4 K sp = 3.5x10 -7 SrF 2 K sp =2.5x10 -9 2) ( 6 pts ) If you dilute an aqueous buffer solution by adding water, will the pH of the solution change, assuming that the Hendeson-Hasselbalch equation can still be applied? ( Circle one) the pH will go down the pH will stay the same the pH will go up Briefly explain your answer. HH is still in effect so the initial concentrations of the acid/conjugate base pair change very little and can still be used to find pH. The volume of the solution changes, but this does not affect the moles of the acid or conj. base so pH does not change.
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