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10 Nov 2019
Consider a solution initially containing 0.500 mol ammonia (NH_3) and 0.300 mol of ammonium ion (NH_4^+). What is the pH after addition of 40 mL of 0.800 M NaOH to thin solution? (NH_4^+, K_a = 5 6 times 10^-10)?
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Irving Heathcote
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Related questions
20 mL of 025 M of NH_3 is titrated with 0.40 M HCl. Calculate the pH of the solution after 20 mL HCl is added. K_b NH_3 = 1.8 times 10^-5 7.00 1.12 12.50 4.74 Beyond the equivalent point, [H_3O^+] from NH_4^+ dissociation is negligible and the pH should be calculated from the excess strong acid. See Section 16.8.
Plusminus Base/Acid Ratios in Buffers Just as pH is the negative logarithm of [H_3O^+], pK_a is the negative logarithm of K_a, pK_a = - log K_a The Henderson-Hasselbalch equation is used to calculate the pH of buffer solutions: PH = pK_a + log [base/acid] Notice that the pH of a buffer has a value close to the pK_a of the acid, differing only by the logarithm of the concentration ratio [base]/[acid]. The Henderson-Hasselbalch equation in terms of pOH and pK_b is similar. POH = pk_b + log[acid]/[base] Part B How many grams of dry NH_4 Cl need to be added to 1.90 L of a 0.500 M solution of ammonia, NH_3, to prepare a buffer solution that has a pH of 8.99? K_b for ammonia is 1.8 times 10^-5. Express your answer with the appropriate units. mass of NH_4Cl = 68.57 g
What is the [H_3 O^+] in a solution that consists of 1.5 M NH_3 and 2.5 NH_4 Cl? K_b = 1.8 times 10^-5 9.3 times 10^-10 M 2.8 times 10^-6 M 1.0 times 10^-5 M 3.0 times 10^-6 M 3.3 times 10^-9 M
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