Calculate the p H of the diprotic acid H2 A (such as H2 SO4) where the dissociation of the first H is strong and the dissociation of the second H is weak but the 5% rule does not apply (K2 is too large relative to the initial concentration of H Aminus). Step 1: H2 A + H2 O – H Aminus + H3 Oplus where K1 is large Step 2: H Aminus + H2 O = A² minus + H3Oplus K2 = 0.012 Assume the initial concentration of H2A is 0.0133 M.

Calculate the p H of the diprotic acid H2 A (such as H2 SO4) where the dissociation of the first H is strong and the dissociation of the second H is weak but the 5% rule does not apply (K2 is too large relative to the initial concentration of H Aminus). Step 1: H2 A + H2 O – H Aminus + H3 Oplus where K1 is large Step 2: H Aminus + H2 O = A² minus + H3Oplus K2 = 0.012 Assume the initial concentration of H2A is 0.0133 M.

Chemistry: An Atoms First Approach

2nd Edition

ISBN:9781305079243

Author:Steven S. Zumdahl, Susan A. Zumdahl

Publisher:Steven S. Zumdahl, Susan A. Zumdahl

Chapter13: Acids And Bases

Section: Chapter Questions

Problem 74E

See similar textbooks

Similar questions

Calculate the p H of the diprotic acid H2 A (such as H2 SO4) where the dissociation of the first H is strong and the dissociation of the second H is weak but the 5% rule does not apply (K2 is too large relative to the initial concentration of H Aminus). Step 1: H2 A + H2 O → H Aminus + H3 Oplus where K1 is large Step 2: H Aminus + H2 O ⇌ A2 minus + H3Oplus K2 = 0.013 Assume the initial concentration of H2A is 0.0107 M. Hint: The H3 Oplus concentration is not zero when setting up your I C E table for step 2 and be sure to consider the total molarity of H3 Oplus when calculating the p H.
Which reaction below will produce significant amounts of product? The acids in decreasing strength are: H 2SO 4, HNO 3, H 3PO 4, HCN HCN + H2PO4− « CN− + H3PO4 H2SO4 + CN− « HSO4− + HCN HNO3 + HSO4− « NO3− + H2SO4 H3PO4 + NO3− « H2PO4− + HNO3
(a) Given that Ka for acetic acid is 1.8 x 10-5 and that forhypochlorous acid is 3.0 x 10-8, which is the stronger acid?(b) Which is the stronger base, the acetate ion or the hypochloriteion? (c) Calculate Kb values for CH3COO- and ClO-.
Write the acid dissociation expression (kb) for the following reaction below: a) NH3 (aq) + H2O(l) <——-> NH4 + (aq) + OH- (aq) Kb = b) Write the conjugate base of H2O, give a brief explanation of how you found the conjugate base of H2O below Please answer both parts otherwise I'll downvote
Why is HCl so acidic? For example, why is H-Cl more acidic than H-S-CH3? Isn't the conjugate base's negative charge more stabilized with a sulfur and extra methyl group (by the induction effect)? While Cl is more electronegative than Sulfur, I would assume Cl being alone as its own ion would make it less stable.
The following scenes represent three weak acids HA(where AX, Y, or Z) dissolved in water (H₂O is not shown):(a) Rank the acids in order of increasing Kₐ. (b) Rank the acidsin order of increasing pKₐ. (c) Rank the conjugate bases in orderof increasing pK(b). (d) What is the percent dissociation of HX?(e) If equimolar amounts of the sodium salts of the acids (NaX,NaY, and NaZ) were dissolved in water, which solution wouldhave the highest pOH? The lowest pH?
9. Order the following solutions according to: a) its increasing acidity: A. pH x 2.00 B. [H3O + ] s 1.00x10 -10 M C. pOH x 1.00 b) His growing basicity A. pH x 2.00 B. pOH x 11.0 C: [H3O+] x 1x10-6 M
answer questions a-c about the Bronsted acid-base reaction below using the identifying letters A-D below each structure. The pKa's for the acids of interest are: acetone(pKa=19.3), and water(pKa=15.7). for parts 1 and 2 part one ( picture 1 )a) The stronger acid is b) Its conjugate base is c) The species that predominate at equilibrium are (two letters, e.g. ac) part two (picture 2) a) The weaker acid is b) Its conjugate base is c) The species that predominate at equilibrium are (two letters, e.g. AC) so which side.
Benozic acid (C6H5COOH) has a pKa of 4.20. a. What is the ratio of the acid to its conjugate base at a pH of 2.20?
Identify which of the acids below will yield the stronger conjugate base and why. a) HOCl (K a = 3.5 x 10 -8 )b) HOI (K a = 2 x 10 -11 )
In calculating pH by ICE method of 0.015 M CH2FCOOH (Ka CH2FCOOH = 2.6 x10-3), the change in CH2FCOOH can not be neglected because. a. Acid Dissociation constant, Ka, has a negative exponent. b. pH is not within +/- pka unit of pka of CH2FCOOH c. initial CH2FCOOH concentration is less than 0.1 M d. dissociation of [CH2FCOOH] will be more than 5% of initial [CH2FCOOH] ----------- To protect corrosion of materials made of iron, the surface of the iron material is electroplated with another metal. Given below information, which metal will be best to “galvanize” the material ? O2 + H2O + 2e- -> 4OH-(aq) .... E0 = 0.401 v Fe2+(g) + 2e- - -> Fe ............ E0 =(-) 0.440 v Cu2+(g) + 2e- - -> Cu ...........E0 = 0.337 v Ni2+(g) + 2e- - -> Ni ............ E0 = (-)0.250 v Zn2+(g) + 2e- - -> Zn ............ E0 =(-) 0.763 v a. Zn b. Ni c. Cu d. Fe2+ ------------ Given : C (s) + H2O(g) <-> CO(g) + H2(g) Which of the following will shift equilibria to form…
Trifluoroethanol (CF3CH2OH, pKa = 12.5) is more acidic than ethanol (CH3CH2OH, pKa = 16). Provide a three-dimensional structure for each conjugate base and draw the orbitals for all lone pairs. Provide a succinct explanation for the greater acidity of trifluoroethanol.