Please select the answer that would complete the following statement corretly. The charge of acetic acid molecules in a 1 molar acetic acid solution is at pH 3.0 and at pH 6.0. Please note the pKa of acetic acid is 3.75 O neutral (0), positive (+1) O positive (+1), neutral (0) O negative (-1), neutra (0) neutral (0), negative (-1)
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- Example The reference range for blood pH is 7.35 – 7.42. What is this range expressed as [H+] in nmol l-1? Calculation: pH = - log [H+] 7.35 = - log [H+] [H+] = antilog -7.35 = 4.47 x 10-8 mol.l-1 = 44.7nmol.l-1 similarly: pH 7.42 = 38.0 nmol.l-1 Range [H+] = 38 – 44.7 nmol.l-1 If the blood pH decreases in an acidosis from 7.42 to 7.15, what is the change in [H+] in nmol.l-1? so this is example on the worksheet, but i still understand how to answer the question.Table 2. Volume of BSA, protein content, and absorbance readings of reference solutions Solution Volume of BSA standard solution (μL) Protein content(μg/mL) Absorbance value At 595 nm 1 0 0 0 2 10 1 0.022 3 30 3 0.065 4 50 5 0.106 5 70 7 0.178 6 100 10 0.299 7 120 12 0.380 Make a graph by plotting the absorbance values versus the BSA protein content (in μg) for theseven reference solutions. When constructing the graph, be…Based on the Henderson-Hasselbalch equation (shown below), calculate the pH when half of a solution of acetic acid is dissociated to acetate (the pKa of acetic acid is 4.76). A. 1.00 B. 3.76 C. 4.76 D. 5.76
- Consider the following pH titration curve of a diprotic acid. What is the approximate values for pka 1 and pka 2? the curve is attached below.The normal range of the sodium electrolyte in the body is 134 to 145 mEq/L. The term mEq is dependent on the charge of the ion. Since sodium is a +1 ion, 134 mEq/L is the same as 134 mmol/L. An ion that has a greater positive or negative charge; however, will have 1 mEq for each positive or negative charge of the ion for every 1 mmol. For example, for Ca+2, 2mEq/1mmol. If the standard range of Magnesium in the body is 0.70 to 0.95 mmol/L, convert this value into mEq/L for the Mg2+ ion. (Use dimensional analysis to figure this out)Three buffers are made by combining a 1M solution of acetic acid and a 1M solution of sodium acetate in the ratios shown in the table below. Which of these statements is true regarding the prepared buffers? (Ka= 1.7x10-5) pH of buffer 1 < pH of buffer 2 < pH of buffer 3 pH of buffer 1 = pH of buffer 2 = pH of buffer 3 pH of buffer 1 > pH of buffer 2 > pH of buffer 3 pH of buffer 1 = pH of buffer 2 > pH of buffer 3 pH of buffer 1 > pH of buffer 2 = pH of buffer 3
- Using this data, calculate the molarity of the oxalic acid standard solution.Define the following:- pH- Buffer- pKaThe pH probe/meter uses following equations: Ecell = L + 0.0592 log a1 = L - 0.0592 pH Where L = L1 + EAg/AgCI + Easy= constants L1 = - 0.0592 log a2 a1 = activity of analyte solution a2 = activity of internal solution How will measured pH value be affected vs “real” pH if HCl in pH electrode, became 0.15M instead of 0.1M ? a.impact can not be determined b.measured pH is higher than "real" pH. c.measured pH is lower than "real" pH. d.measured pH is same as "real" pH.
- Give full explanation The physcian orders ascorbic acid 0.25mg IM for your patient admitted with an alcohol problem. You have ascorbic acid 500mg/mL. How many milliliters will you administer?Given the following information, calculate the total activity in the undiluted protein sample. Activity of 1 ml of diluted sample = 0.5 Total volume of sample = 5 ml Dilution factor = 10 25 50.5 250 2.5You are supplied with the following: NaCl (Mr = 58.443 g/mol) 2.5 M Tris-Cl, pH 8 oplossing / solution (1 Litre) EDTA, sodium salt (Mr = 380.2 g/mol) 10 % Sodium dodecyl sulfate solution Proteinase K solution (50 mg dissolved in 1 ml ddH2O)You need a digestion buffer consisting of the following: 15 mM NaCl 75 mM Tris-Cl, pH 8 16 mM EDTA, pH 8 0.8 % Sodium dodecyl sulfate 0.75 mg/ml proteinase KCalculate: How will you prepare 500 ml of the digestion buffer?