A bottle contains an aqueous solution of sulfuric acid of pH 4.5. The hydroxyl i

Question

A bottle contains an aqueous solution of sulfuric acid of pH 4.5. The hydroxyl ion (OH^-) concentration of such a solution is (a) 3.22 times 10^-10 M (b) 4.22 times 10^-11 (c) 5.22 times 10^-12 (d) 6.22 times 10^-13 (e) None of the above The H^+ ion concentration, pH, OH^+ ion concentration and pOH of a 0.001 M solution of Ca(OH)_2 are (a) 0.02M, 2.5, 10^-3 M and 11 respectively (b) 0.002M, 2, 10^-7 M and 9 respectively (c) 0.003M, 2.2, 10 M and 8 respectively (d) 0.0005 M, 2.3, 10^-9 M and 14 respectively (e) None of the above. A segment of a DNA chain with the base sequence -^5'TCAGAT^3'- would be transcribed by RNA polymerase to an RNA chain with which of the following base sequences? (a) CUAAGU (b) AGUCUA (c) AGTCTA (d) AUCUGA (e) ATCTGA. Which of the following ionic species of glutamic acid would be most prevalent at pH 7? (a) HOOC-CN^+H_3-H-(CH_2)_2-COOH (b) 'OOC-CN^+H_3-H-(CH_2)_2-COOH (c) 'OOC-CN^+H_3-H-(CH_2)_2-COO' (d) 'OOC-CNH_2-H-(CH_2)_2-COO' In biological membranes, proteins and lipids interact mainly by (a) Hydrophobic interactions (b) Hydrogen bonding (c) Hydrophilic interactions (d) Ionic bonds. Sucrose is a disaccharide, which is made up of (a) glucose and sucrose (b) glucose and fructose (c) glucose and glucose (d) fructose and fructose (e) glucose and galactose. Peptide bond synthesis requires an input of energy during (a) Amino acid activation (b) The binding of the aminoacyl-tRNA to the A site on the ribosome (c) The movement of the peptide-tRNA to the P site and the associated movement of the mRNA (d) All of the above. At pH 7.0, the net charge on the peptide Ala-Gly-Glu-Pro- would be (a) -2 (b) -1 (c) 0 (d) 1 (e) +2. Name all the amino acids with basic side chains:

Explanation / Answer

1)
pH = 4.5
pOH = 14 - pH
= 14 - 4.5
= 9.5

use:
pOH = -log [OH-]
9.5 = -log [OH-]
[OH-] = 3.17*10^-10 M

Answer: a

2)
1 mol of Ca(OH)2 has 2 mol of OH-
[OH-] = 2*[Ca(OH)2] = 2*0.001 M = 0.002 M

[H+] = (1.0*10^-14) / [OH-]
= (1.0*10^-14) / (0.002)
= 5.0*10^-12 M

pH = -log [H+]
pH = -log (5.0*10^-12)
pH= 11.3

pOH = 14 - pH
= 14 - 11.3
= 2.7

Answer: e

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