lab 5 solubility prduct wk 2

1. Molarity of the NaOH solution 0.238 mol/L Trial 1 Trial 2 Trial 2. Volume of H;PO4 added to flask 22.0 mL 22.0 mL 22.0 mL 3. Initial NaOH volume _0.25_ mL _0.75_ mL 0.55 mL 4. Final NaOH volume 17.60_ mL 18.20_ mL _17.90_ mL 5. NAOH volume used for titration to reach green end point mL mL mL 6. NaOH volume used for titration 7. Moles of NaOH used for titration mol mel mel 8. Moles of H;PO4 that reacted mol mel mol 9. Volume of H;PO4 added to flask L 10. Molarity of H3PO4 mol/L mol/L mol/L

If you could help with the below pre-lab questions please. PRELABORATORY ASSIGNMENT Define the term "salt.'  The solubility of AgCl(s) in water is 1.3 x 10-5 M.  Calculate its Ksp.  The solubility of CsF2(s) in water is 2.1 x 10-4 M.  Calculate its Ksp. The of Pb(OH)2 is 4 x 10-15.  Calculate the molar solubility of this compound.

5. From the stock solutions of 1.00-mol L acetic acid and 1.00-mol L of sodium acetate, six solutions were prepared with the acetic acid and acetate concentrations shown in the table below. The pk of acetic acid is 4.756. [Acetic acid). [Acetate). Final volume, Solutio mol L mol L ml A. 0.10 100 0.0000351 100 0.10 0.10 100 0.010 0.010 100 0.10 0.15 100 0.15 0.10 100 -1 Determine the volume of the stock solutions of 1.00-mol Lacetic acid and 1.00-mol L of sodium acetate needed to prepare the solutions. Write your answer in the table below. a. Vol. acetic acid, ml Vol. of acetate, ml Solution A b. Calculate for the initial pH of the solutions and write the pH in the table below. Solution Initial pH A Which solution(s) have the same pH? C.

B. Molar Concentration of an Acid Solution Acid type, HA or H,A: HA Unknown No. Balanced equation for neutralization of acid with NaOH. NaOH + HCI = NaCI + Sample 1 Sample 2 Sample 3 1. Volume of acid solution (mL) 25.0 25.0 25.0 2. Buret reading of NaOH, initial (mL) 0.00 0.00 0.00 3. Buret reading of NaOH, final (mL) 26.3 26.0 26.8 26.3 26.0 0.148 4. Volume of NaOH dispensed (mL) 26.8 5. Molar concentration of NAOH (mol/L), Part A 6. Moles of NaOH dispensed (mol) 7. Molar concentration of acid solution (mol/L) 8. Average molar concentration of acid solution (mol/L) Data Analysis, B 9. Standard deviation of molar concentration Data Analysis, C 10. Relative standard deviation of molar concentration (%RSD) Data Analysis, D

6-20. A 4.476-g sample of a petroleum product was burned in a tube furnace, and the SO, produced was col- lected in 3% H,O2. Reaction: SO-(g) + H;O, → H,SO4 A 25.00-mL portion of 0.00923 M NaOH was intro- duced into the solution of H,SO,, following which the excess base was back-titrated with 13.33 mL of 0.01007 M HCI. Calculate the parts per million of sulfur in the sample.

Part A) A 10.0 mL sample of 0.250 mol/L NiF2(aq) is mixed with 20.0 mL of 0.0900 mol/L NaOH(aq) and then diluted to a final volume of 100. mL.Calculate the concentration of Ni2+ ions in the 100 mL mixture before the reaction starts. Express your answer to three significant figures.ksp: 5.48e-16

9. To correctly adjust the pH of a solution to be used in an electrophoretic separation procedure, exactly 60.81 mL of 1.209 M NaOH were added to 100.00 mL of 1.418 M lactic acid (HLac). HLac + NaOH →NaLac + H₂O Please answer the following questions. A. B. C. Determine the equilibrium concentration of lactic acid in the buffer solution. Please provider your answer below. 0² 0₂ Determine the equilibrium concentration of NaLac in the buffer solution. Please provider your answer below. 7² 0² $ → → S ■ Given that Ka = 1.403×10-4 for lactic acid, determine the pH of the buffer solution. pH = Please provider your answer below. M A $ ■ M

Why does it conduct more before and after this minimum point?

If 2.00 ml of 0.200M NAOH are added to 1.00 L of 0.100M CaCl2, will a precipitate form? Ksp Ca(OH)2= 8.0 x 10-6 Sample answer format: 1.0x10-14(YES) (Note: Case, Space, and format sensitive)

up the 1. In a reaction involving the iodination of acetone, the following volumes were used to make reaction mixture: 10 mL 4.0 M acetone + 10mL 1.0 M HCl + 10 mL 0.0050 M 1₂ + 20 mL H₂O a. How many moles of acetone were in the reaction mixture? Recall that, for a component A, no. moles A = MA X V, where MA is the molarity of A and V is the volume in liters of the solution of A that was used. b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50 mL, 0.050 L, and the number of moles of acetone was found in Part a. Again, MA = moles acetone no. moles A V of soln. in liters M acetone

Would the last answer be right? Thanks!

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Determine concentration of OH- in a 0.724 M solution of BrO- (Kb = 4.0 × 10-6). 1 2 NEXT > Based on the given values, set up ICE table in order to determine the unknown. BrO (aq) H:O(1) OH-(aq) HBRO(aq) + + Initial (M) Change (M) Equilibrium (M) 5 RESET 0.724 4.0 x 10-6 0.362 +x -X +2x -2x 0.724 + x 0.724 - x 0.724 + 2x 0.724 - 2x 4.0 x 106 + x 4.0 x 106 - x 4.0 x 10-8 + 2x 4.0 × 10-8 - 2x

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14) Use 4-5 complete sentences to clearly explain how it is possible for a reaction to have a positive value of AG n but a negative value for AG. page 3 of 3 Scanned with CamScanner

Give the expression for the solubility product constant for Ca3(PO4)2- Select one: O a. B Ca2+PR PO43-P Ca3(PO4)2 O b. [ Ca2+P[ PO43-P Ca3(PO4)2 O c. R Ca2+P[3 PO43-3 Ca3(PO4)2 O d. [ Ca2+j?[ PO43-73 O e. [Ca2+j³[PO43-12

Why does it not conduct at this low point?

The buret was filled with 0.100 M HCl solution. Then was transferred in a 25.0 mL of saturated calcium hydroxide solution (2g of calcium hydroxide per 100 ml of water) in two separate E-flasks. Then 2 drops of phenolphthalein was added to each flask Titration data for the determination of solubility and Ksp of calcium hydroxide: Trial 1: Final Buret reading (ml)-19.80; Initial Buret reading (ml)- 13.00; Temperature (Celcius)- 25 Trial 2: Final Buret reading (ml)-26.10; Initial Buret reading (ml)- 19.80; Temperature (Celcius)- 25 Voume of HCl used: Trial 1- 6.80ml; Trial 2- 6.30mL 1. Compute for the moles of H+ used and the moles of OH- present. moles of H+ used = (concentration of HCl) × (volume of HCl used)moles of OH- = moles of H+ used 2. Construct an ICE table for the reaction.3. Calculate the molar solubility (in mol/L) of OH- and Ca2+.4. Determine the solubility of Ca(OH)2 in g/L. (MM of Ca(OH)2 = 74.096 g/mol).                                                          5.…

1. In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture. 1.0 mL 4.0 M acetone + 1.0 mL 1.0 M HCl + 1.0 mL 0.0050 M I₂ + 2.0 mL H₂O a. Calculate the initial (diluted) concentration of acetone in the reaction mixture. [acetone]ini = = b. Calculate the initial (diluted) concentration of the hydrogen ion, H+, in the reaction mixture. [H*]ini c. Calculate the initial (diluted) concentration of iodine, I2, in the reaction mixture. [12]ini = =

For the reaction: PbCl2(s) Pb2+(aq) +2C11-(aq), what is Q* when 8.0 mL of 0.050 M lead nitrate is added to 14 mL of 0.031 M sodium chloride? Ksp of lead chloride is 1.6 x 10-5 M³. Hint given in general feedback *Recall: Q is compared to Ksp to determine whether a precipitate forms.

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Table 1: Absorbance data recorded for five 0.200 M Fe(NO:)3 solutions at 20.0 °C Beaker 0.0020 M [FESCN2+] (M) Abs. 0.00200 SCN (mL) Fe(NO3)3 (mL) 5.0 0.00 0.00 M -0.001 2 5.0 2.00 8.0 x 10-5 M 0.278 5.0 3.00 1.2 x 10-4 M 0.415 4 5.0 4.00 1.6 x 10-4 M 0.555 5.0 5.00 2.0 x 10-4 M 0.671 Part 1: y=3386x+0.0043 R^2=0.9991 Part II – Determining Concentration of an Unknown Table 2: Best-fit line and Absorbance data of unknown solution Best-fit line equation from Part I Absorbance reading of unknown 0.338 Part III – Determining the Equilibrium Constant Test tube 0.0020 M 0.00200 H20 (mL) Abs. Net Abs Fe(NO3)3 (mL) SCN-(mL) 1 3.00 0.00 7.00 0.000 2 3.00 2.00 5.00 0.119 0.119 3 3.00 3.00 4.00 0.212 0.212 4 3.00 4.00 3.00 0.265 0.265 5 3.00 5.00 2.00 0.335 0.335

Use the References to access important values if needed for this question. The Solubility Product Constant for lead phosphate is 3.0 x 10-44. Solid sodium phosphate is slowly added to 75.0 mL of a 0.0367 M lead acetate solution. The concentration of phosphate ion required to just initiate precipitation is M. 20 C Submit Answer $ 4 R FI V Retry Entire Group 9 more group attempts remaining % 5 T Cengage Learning | Cengage Technical Support G ^ 6 B MacBook Pro Y H & 7 U N * 00 8 J I M ( 9 K O ) -O 4 L | ' P Previous Next> Email Instructor + = [ Save and Exit 21 I 1 c

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Determining the Ksp of Calcium Hydroxide: Ca(OH)2 (s) ↔ Ca2+ (aq) + 2OH– (aq) Ksp = [Ca2+][OH–]2 The molarity of the hydrochloric acid used was 0.050 M.15.00 mL of filtered Ca(OH) 2 was measured out for the titration.Calculate the [Ca 2+ ]. Explain, please.  Trial Initial Buret Reading  Endpoint Buret Reading  Endpoint Volume  Equivalence Point Volume  1 80 ml 9.31 pH 12 ml 12.60ml

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