A monoprotic weak acid (HA) has a pKa value of 4.896. Calculate the fraction of HA in each of its forms (HA, A¯) at pH 6.262. -9 -8 28.34 ×10 2.834 ×10 CHA = CA- = [A"] What is the quotient at pH 6.262? [HA] [A"] [HA] 21.17
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![A monoprotic weak acid (HA) has a pKa value of 4.896. Calculate the fraction of HA in each of its forms (HA, A¯) at pH 6.262.
28.34 x10
-8
2.834 x10
AHA =
dA- =
What is the quotient
[A]
at pH 6.262?
[HA]
[A*]
[HA]
21.17](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf8dd5bb-1368-4a10-9929-9294802d0b74%2Fb7aa0cf2-24b4-46a4-a46b-e00bb1811874%2F8092bhn_processed.png&w=3840&q=75)
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- Calculate the pH of each of the following solutions. (a) 10.0 mL of 0.300 M hydrofluoric acid plus 30.0 mL of 0.100 M sodium hydroxide (b) 100.0 mL of 0.250 M ammonia plus 50.0 mL of 0.100 M hydrochloric acid (c) 25.0 mL of 0.200 M sulfuric acid plus 50.0 mL of 0.400 M sodium hydroxideA monoprotic weak acid (HA) has a pKa value of 4.713. Calculate the fraction of HA in each of its forms (HA, A-) at pH 5.848. aHa= Aa- What is the quotient (a-/ha)= at ph 5.848 ?From the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb for each of the weak bases. (H3O+) = 1.34 × 10−3 M;(CH3CO2−) = 1.34 × 10−3 M;(CH3CO2H) = 9.866 × 10−2 M; (OH−) = 4.0 ×10−4 M;(HClO) = 2.38 × 10−4 M;(ClO−) = 0.273 M; (C6H5NH3+) = 0.233 M;(C6H5NH2) = 2.3 × 10−3 M;(H3O+) = 2.3 × 10−3 M
- Is the pKa for the acid in the question below 3.75 or 4.75. If the pKa is wrong, the whole answer is wrong. https://www.bartleby.com/questions-and-answers/what-is-the-weight-nach3coo-of-that-must-be-added-to-750.0-ml-of-0.180-m-acetic-acid-to-form-a-ph-5./e13e02da-a2a6-4137-b10a-cf4e549d8584 The original question is: What is the weight NaCH3COO of that must be added to 750.0 mL of 0.180 M acetic acid to form a pH = 5.25?9. Diethyl amine, (CH3)2NH, has a Kb of 5.81x10-4 at 25 Ca. Write the dissociation reactionb. Calculate the pH of a 0.060 M solution c. Calculate Ka from Kb 10. The phosphate ion, PO4-3, acts as a Bronsted base in aqueous solutions. The Kb for phosphate at 25 C is 2.07x10-2a. Write the base reaction between phosphate ion and waterb. Calculate the pH of a 1.2 M solution 11. The carbonate ion, CO3-2, acts as a Bronsted base in aqueous solutions. The Kb for carbonate at 25 C is 2.19x10-4a. Write the base reaction between carbonate ion and waterb. Calculate the pH of a 0.15 M solution1. Arrange the following in order of increasing pH of their corresponding solutions. a. CH3NH3Cl, NaCH3COO, RbBr b. K2C2O4, C2H5NH3Br, Ba(ClO4)2 2. Oleic acid (which can be denoted as HOL), is a weak acid. Considering its sodium salt, sodium oleate (NaOL), calculate the pKa of HOL given that a 0.100 M solution of NaOL has a pH of 9.01. (hint: Kw=Ka * Kb).
- From the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb for each of the weak bases.(a) NH3: [OH−] = 3.1 × 10−3 M;[NH4+] = 3.1 × 10−3 M;[NH3] = 0.533 M;(b) HNO2: [H3 O+] = 0.011 M;[NO2−] = 0.0438 M;[HNO2] = 1.07 M;(c) (CH3)3N: [(CH3)3N] = 0.25 M;[(CH3)3NH+] = 4.3 × 10−3 M;[OH−] = 4.3 × 10−3 M;(d) NH4 + : [NH4 +] = 0.100 M;[NH3] = 7.5 × 10−6 M;[H3O+] = 7.5 × 10−6 MFrom the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb for each of the weak bases.(a) CH3CO2H: [H3 O+] = 1.34 × 10−3 M;[CH3 CO2−] = 1.34 × 10−3 M;[CH3CO2H] = 9.866 × 10−2 M;(b) ClO−: [OH−] = 4.0 × 10−4 M;[HClO] = 2.38 × 10−5 M;[ClO−] = 0.273 M;(c) HCO2H: [HCO2H] = 0.524 M;[H3 O+] = 9.8 × 10−3 M;[HCO2−] = 9.8 × 10−3 M;(d) C6 H5 NH3+ : [C6 H5 NH3+] = 0.233 M;[C6H5NH2] = 2.3 × 10−3 M;[H3 O+] = 2.3 × 10−3 MSaccharin is a monoprotic acid. If the pH of a 6.30 x 10–3 M solution of this acid is 5.63, what is the Ka of saccharin? explain in detail what is this question about.
- Which of the following sequences show an increasing strength of acidity? I. HNO2 (Ka=4.0 x 10-4) II. HOC6H5 (Ka=1.6 x 10-10) III. HSO4- (Ka=1.2 x 10-2) IV. HCN (Ka= 6.2 x 10-10) III, I, IV, II II, IV, I, III III, I, II, IV IV, II, I, IIIConsider the molecule, pentanoic acid which has a pKa of 4.81: H3C OH Calculate the conjugate base to acid ratio at each of the following pH values. i. 11.75 ii. 4.81 iii. 2.955,Calculate the [OH-]of the solution with pH=10.50 Group of answer choices 1.05 x 10-10 5.0 x 10-10 3.2 x 10-10 3.2 x 10-4 6, Use the Ka values for the following acids to find the strongest conjugate base: HSO4- 1.2 x 10-2H2PO4- 6.3 x 10-8HCO3- 4.7 x 10-11 Group of answer choices A, H3PO4 B, H2SO4 C, HPO42- D, H2CO3 E, CO32- F, SO42-
