580360 Isolation of Casein Q

pH pH = 4.389 pH = 4.874 = pH = 5.339 pKa Answer Bank Using the equation to calculate the quotient [A]/[HA] at three different pH values. [A] [HA] + log [A-] [HA] [A] [HA] || = H = [A] [HA] 1 0.3273

5

PEW41L216q47fR9EANXYnmJnqrAsUafoA6oj JhzTEL52w13Gy9w6bzhg%253D%253D?gameType=async This citrus fruit commonly consumed at home has a ph level of F7 DELL F8 F9 above 7 0 F10 F11 O F12 Exit session 2309 3139 [] Back Oct 26 11:18 0 PrtScr Insert (p Next Delete

09. Two samples of the same soil is subjected to pH analysis using a pH meter, the first is analysed using 1:1 soil:water ratio and the other with 1:1 soil:KCl ratio, what would be the most probable result?        A. Soil analused using 1:1 soil:KCl ratio will register a higher pH compared to that of soil analysed using 1:1 soil water ratio       B. Soil analysed using 1:1 soil:KCl will register a lower pH compared to that of soil analysed using 1:1 soil:water ratio       C. Soil analysed using 1:1 soil:water ratio will register a lower pH compared to that of soil analysed using 1:1 soil:KCl ratio       D. Soil analysed using 1:1 soil:water ratio will have the same pH with that of soil analysed using 1:1 soil:KCl ratio

Can you please show me the computation or how to compute? Thank you.

the question states that the Ka value was 3.0 x 10-8 but in step two to calculate the value of pKa you calculated using the -log of 30.0 x 10-8 why did you use that number instead?

Data in first picture and questjons in second

Complete the following table "predicted pH." Please show all work on a separate sheet of paper and attach. Note that the pK, of H₂PO4 is 7.2. Actual pH Predicted pH Original After 45 µL of 5 M NaOH solution 4.30 6.05 After 205 μL more of 5M NaOH 6.90 After 205 μL more of 5M NaOH 11.01 After 45 μL more of 5 M NaOH 11.29

A weak base has KB = 3.26 x 10-2. What is the pKB of this base? Report your answer to at least 1 decimal place.

Electrical Conductivity of solution lons released in solution Solution pH Taste 1 Very high 8.0 bitter ОН- 2 Very high 14.0 bitter ОН - none 7.0 none both 4 Very high 2.0 H+ sour Based on the experimental data collected by a student above, which solution is acidic? Give at least 2 reasons to support your answer.

Use the "Square of Death" to determine the following for the solution in problem 6 where pH = 11.3010 [OH-¹] = [H3O+¹] = pOH = -log [OH-¹] = a. acetic b. hydrochloric g. sulfurous f. sulfuric c. phosphoric d. perchloric h. hydrosulfuric i. OH-1 j. H30+1 k. SO4² 0. H₂PO3-1 p. CO3² q. HCO3-¹ e. hydrofluoric 1. Mg+2 -2 -2 n. HPO3-² -2 r. H₂S t. S-² u. 0.2500 v. 0.1000 S. HS-1 y. 3.0000 w. 0.02500 z. 1x 10-14 x. 1.000 x 10-3 dd. 2 aa. 2.6990 ee. 1 ff. 2.000 x 10-³ gg. 5.000 x 10-12 mm. conjugate acid bb. 1.000 x 10-11 hh. 11.3010 oo. base CC. 11.0000 ii. 0.01250 pp. conjugate base jj. 0.22100 kk. 0.05656 m. Cl-1 II. acid

Equation 10-12: pH = ½ (pKa₁ + pKa2) is applicable for which of the following solutions? I. 1.0 x 10-3 M H3PO4 pka3 = 12.375] II. 1.0 x 10-3 M H2SO3 III. 1.0 x 108 M H2SO3 IV. 1.0 x 10-8 M H2C2O4 [pKa₁ = 2.148; pKa2 = 7.198; [pKa₁ = 1.857; pKa2 = 7.172] [pKa₁ = 1.857; pKa2 = 7.172] [pKa₁ = 1.250; pKa2 = 4.266] a. Il only b. II and IV c. IV only d. I only e. I, II, III, and IV f. Ill only g. I, II, and IV The pH of 0.050 M K₂HPO4 is Use the pKa values given in Question 9. The pH of 0.050 M K3PO4 is Use the information given in Question 9 for H3PO4. Hint: A quadratic equation is involved.

The pH of 0.010 M trimethylamine is 10.88. What is the value of Kb for this base? 1.3 * 10-11 9.8 * 10-8 4.8 * 10-7 5.8 * 10-5 7.6 * 10-4

2. Determining the Acid Content of Vinegar Trials A В C Volume of Vinegar used (ml) 25.00 diluted to 250.00 mL Volume of Aliquot used (ml) 50.00 50.00 50.00 Final Volume Reading NaOH (ml) 36.25 35.10 36.94 Initial Volume Reading NaOH (ml) 0.00 0.10 0.20 Net Volume NAOH used (ml) Calculation of % (w/v) CH3COOH via dimensional analysis Trial A Trial B Trial C Calculation of Average percent (w/v) CH;COOH

5. Value for Kw at 50.00°C is 5.470 x 10-14. Calculate the pKwat 50.00°C. 6. Value for Kw at 100.0°C is 4.900 x 10-13. Calculate the pH of a 1.000 x 10 -2M NaOH solution at 100.0°C.

1. Prepare 50-mL of a 50 mM H,PO, solution in a 50-ml conical tube. Molar mass of NaH2PO4 H2O used is 138 moles 2. Check the pH of this solution and record. 3. A5 M NAOH solution has been provided. 4. Add 45 µl of 5 M NaOH into the 50 mM H2PO, solution. Check the pH and record. 5. Add 205 µl of 5M NaOH into the 50 mM H2PO4 solution. Check the pH and record. 6. Add 205 µL more of 5M NaOH into the 50 mM H2PO, solution. Check the pH and record. 7. Add 45 µl of 5 M NaOH into the 50 mM H2PO, solution. Check the pH and record. Complete the following table "predicted pH." Please show all work on a separate sheet of paper and attach. Note that the pK, of H2PO, is 7.2. What would be the predicted pH's? Would they be similar to the actual pH's from procedure? After 45 µl of 5 | After 205 µl more of 5M NaOH Original After 205 µl more After 45 µl more solution M NaOH of 5M NaOH of 5 M NaOH Actual pH 4.30 6.05 6.90 11.01 11.29 Predicted pH

At 10.ºC, Kw = 1.14 x 10-15.  What is the pH of water at 10.ºC?  Choose the closest answer. 3.98 4.56 6.15 7.00 7.47

Part A Calculate the pH of a solution containing a caffeine concentration of 415 mg/L. Express your answer to one decimal place. V ΑΣΦ ΑΣφ ? pH = %3D

e proton concentrations of three solutions at 25 °C are given Basic Acidic Neutral Answer Bank [H]= 1.0 x 10 ° M H]=0.00001 M [H']= 1.0 x 10 ' M about us terms of use privacy policy careers conta 996+ DELL

Rank the following in order of increasing pH.  Explain the basis of your ranking.  Assume the concentrations are all 0.10 M. Attached Table Below. Piperazine Pyridine hydrochloride Phenylalanine Disodium phthalate Sodium prolinate Phosphoric acid.