Will a ppt of A82CO3 (s) form if you mix 3.9 x 10-3 M Ag* (aq) with 1.7 x 10-4M CO32- (aq)? + (Which of the following is entirely correct?) Q = 6.6 x 10', no ppt Q = 2.6 x 10, no ppt OQ = 2.6 x 10°, ppt forms %D Q = 1.3 x 106, ppt forms %D Q = 6.6 x 10/, ppt forms
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- Which of the following is more likely to precipitate the hydroxide ion? Cu(OH)2(s) *=» Ctr+(aq) + 2 OH’(aq) K = 1.6 X IO-19 Ca(OH)2(s) *=» Ca2+(aq) + 2 OH’(aq) K = 7.9 X 10"6Will a ppt of PbSO4 (s) form if you mix 7.6 x 104 M Pb(NO3)2 (aq) with 9.3 x 10-2 M Na2SO4 (aq)? (Which of the following is entirely correct?) Q = 7.1 x 10³, no ppt Q = 1.3 x 108, no ppt %D Q 5.0х 10%, nо pt Q 5.0 x 10, ppt forms %| Q = 7.1 x 105, ppt formsThe weak monoprotic acid, acetic acid, is titrated with the strong base, potassium hydroxide as follows: HC2H3O2(aq) + K+ OH- (aq) → K+ C2H3O2-(aq) + H2O(l) Ka for acetic acid is 1.81 x 10-5 (at 25 oC). A 25.00 mL sample of a solution of acetic acid with concentration 0.0833 M is titrated with 0.1000 M KOH. a) Sketch the pH vs volume of added base titration curve for this reaction. pH vertically, volume of base horizontally. Label your axes with correct pH and volumes. (Attach more space, if needed) b) what is the pH at the beginning of the titration, Vbase = 0.00 mL ? c) what is the volume of the base needed to reach the equivalence point ? (label the equiv. point) d) what is the pH at the equivalence point?e) what is the pH of the titration when 5.00 mL of base have been added ? f) what is the pH when the volume of base added equals half the volume of the equivalence point? g) what is the pH of the titration when 20.00 mL of base have been added? h) what is the pH of the titration…
- The weak monoprotic acid, acetic acid, is titrated with the strong base, potassium hydroxide as follows: HC2H3O2(aq) + K+ OH- (aq) → K+ C2H3O2-(aq) + H2O(l) Ka for acetic acid is 1.81 x 10-5 (at 25 oC). A 25.00 mL sample of a solution of acetic acid with concentration 0.0833 M is titrated with 0.1000 M KOH. 1. what is the pH when the volume of base added equals half the volume of the equivalence point? 2. what is the pH of the titration when 20.00 mL of base have been added? 3. what is the pH of the titration when 30.00 mL of base have been added?An aqueous solution contains a mixture of 0.0026 M of Fe 2+ and 0.0026 M of Al3+ ions. In order to separate these ions, an aqueous solution of sodium hydroxide has to to be added. Given that Ksp of Fe(OH)2 is 8.0 X 10-16 and the Ksp of Al(OH)3 is 4.6 X 10-33 what is the concentration of the sodium hydroxide needed to precipitate out the first cation?A 0.3749 g soda ash sample is analyzed by titrating sodium carbonate with the standard 0.2388M HCl solution, requiring 49.38ml. The reaction is: CO32- + 2H+ → H2O +CO2Calculate the percent sodium carbonate in the sample.
- The weak monoprotic acid, acetic acid, is titrated with the strong base, potassium hydroxide as follows: HC2H3O2(aq) + K+ OH- (aq) → K+ C2H3O2-(aq) + H2O(l) Ka for acetic acid is 1.81 x 10-5 (at 25 oC). A 25.00 mL sample of a solution of acetic acid with concentration 0.0833 M is titrated with 0.1000 M KOH. A. What is the pH at the beginning of the titration, Vbase = 0.00 mL? B. What is the pH at the equivalence point? C. What is the pH of the titration when 5.00 mL of base have been added? D. What is the pH when the volume of base added equals half the volume of the equivalence point? E. What is the pH of the titration when 20.00 mL of base have been added? F. What is the pH of the titration when 30.00 mL of base have been added?Will a ppt of Au(OH)3 (s) form if you mix 8.3 x 10-24 M Au3+ (aq) with 6.3 x 10-13 M OH (aq)? (Which of the following is entirely correct?) Q = 2.1 x 10-60, no ppt Q 1.6 x 10 35, ppt forms Q 5.2 x 10-36, no ppt Q = 5.2 x 10-36, ppt forms 2.1 x 10-60, ppt formsThe concentration of an aqueous solution of CaSO4 (Ksp = 7.1 × 10–5) is 8.43 × 10–3 M. Qsp = ___ and this solution is ___.
- Chemistry If 110.0 mL of 0.0030 MNa2SO4(aq) is saturated with CaSO4, how many grams of CaSO4 would be present in the solution? [Hint: Does the usual simplifying assumption hold?] Express your answer using two significant figures. This answer not 0.07 0.084, 0.09You added 0.05 mL of 0.10 M AgNOs to 4.0 ml of 2.0 M NaCI, and formed a saturated solution of AgCI (s):AgCI (s) 2Ag (ag)+ CI (aq)К = КэрUsing this saturated solution as the test solution you set up the following cellAg|Ag (test solution after reaction)llAg (1.0M)Agand measured Ecell = 0.61 V.(a) Using the measured value of Eceit and Eq (15), calculate the equilibrium [Ag ] (i.e. of the test solution after reaction).Molarity of the standard solution of sodium hydroxide = 0.1010 mol/L %3D Volumes of NAOH(aq): Third Rough titration First Second titration titration titration Final burette reading: 18.48 33.58 48.40 23.72 (mL) Initial burette reading: (mL) Volume of NaOH(aq) Titrated (added) (mL) 3.48 18.48 33.60 8.73 Fourth Fifth titration titration Final burette reading: (mL) Initial burette reading: (mL) Volume of NaOH(aq) Titrated (added) (mL) 38.68 20.62 23.80 5.75 4. (a) The sodium ion did not take part in this chemical reaction. What do we call such an ion? (b) Draw a simple diagram which shows how the sodium ion mixes with water in solution. What do we call this physical process? 4
