answer question 29 show all work containing 1oO of the folowin solutions amistur
ID: 697398 • Letter: A
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answer question 29 show all work
containing 1oO of the folowin solutions amisture comain npare the ssociation of the acil in Exercise 17a with the acid in Exercise 17d. Explain the The cetcevvltom MCL h Caleolate the phh wte this solution Negle antly ent dissociation of the ncid ent dissoctation of ionization of the base in Exercise 18a with een of the base in Exercise 18d. Explain any 37. Calculate tve mass ent 500 0 ml of .200 olution e 0.020 mol HCI is added to 1.00 L of each 0.020 mol HCl is added to 1.00 Lof each tier 0.020 mol NaOH is added to 1.00 L. of after 0.020 mol NaOH is added to 1.00 L of 3H. What volves o the pH after after mixed to prepare olutions in Exercise 17 Calculate the pH ane of the four 39. Consider a solu Calculate the ra following pH a PH 4.50 b. pH5.0 40. Calculate th atslutions in Exercise 18 n of th ulate the pH after ur solutions in Exercise 17 solutions h of the cul 0.020 ons in Exercise 17 shows the least change in Calculate each or t late the pH 24, tch of the solutions in Which of the solutio pft u dition of acid or base? Explain. unhe solutions in Exercise 18 is a buffered solution? the 41. Conside choice 10 L the pH of a solution that is 1.00 M HNO2 and 1.00 M 42. Con cho DH of a solution that is 0.60 M HF and 1.00 M KF H after 0.10 mol NaOH is added to 1.00 L of the Exercise 27, and calculate the pH after 0.20 mol HC NaNO, 43. C in Exe solution to 1.00 L of the solution in Exercise 27Explanation / Answer
For HNO2, pKa = 3.35
1 L of the solution in part (27) contains 1 mole of acid HNO2 and 1 mole of salt NaNO2.
When 0.1 moles of NaOH is added, moles of acid decrease and moles of salt increase by this amount.
Using Henderson Hasselbach equation:
pH = pKa + log(moles of salt/moles of acid)
Putting values after addition of 0.1 moles of base NaOH:
pH = 3.35 + log((1+0.1)/(1-0.1)) = 3.437
When 0.2 moles of HCl is added, moles of acid increase and moles of salt decrease by this amount.
Using Henderson Hasselbach equation:
pH = pKa + log(moles of salt/moles of acid)
Putting values after addition of 0.2 moles of acid HCl:
pH = 3.35 + log((1-0.2)/(1+0.2)) = 3.174
Hope this helps !
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