Question

4. A solution is made by dissolving 15.00 grams of acetic acid and 3.25 grams of sodium acetate in 500.00 mL of water. What is the pH of the solution? The K, of acetic acid is 1.8 x 10. (Hints: First convert grams to mol and then determine the molarity of acetic acid and sodium acetate, ...) 5. What is the pH of a 2.50 molar solution of NHCl(aq)? The Ko for NH, is 1.8 x 105. 6. A certain acid has a pKa equal to 4.88. Is this acid stronger or weaker than acetic acid, which has a pka of 4.74? What is the value of Ka for this acid? 7. In a sample of blood at 25*C(H) = 4.6 x 10* M. Find the molar concentration of OH, and decide if the sample is acidic, basic, or neutral.

Answer 1

The solution of all the parts is given below with explanation.

Here So we have mans of Acetic acid = 15 g. 60.059 Molar man of acetic acid- moles of acetic acid mass molar non Thus no 15 -0.249 note 60.05 Of Aus molimon Now, holes of sodium acetale = given maar 3.25 = 0.039 mole 82.03 The ph of a weak acid with strong base is given by Now, done Nhu ان لاله 2. pH = pka + log [ Base [Acid] Here we have to for actic acid 1.800* pka = -log (ka) - log (1.8x105) 4.74. Thus ph= 4.74 + log [GYOONa] [choon] Molarity of Sodium acetate = moles- Now volume of solution 0.039 0.078 mollh Molarity ob Arete ood s 0.249 = = 0.498 ml 5.5

This pH = 4.79 + log [0 078] pH = 3.93 50.4987 We have given We know that, kaxky - 14 Os Aus Kan for NH₂ is 18410 1x 10" ka - 1x 16 Kb 1x10-14 1.8x105 ka 5.55* 10 9 Now, the hydrolysis of Nilty Cl will be done Nilz + H₂ 2.5 Nhy Cloog? + H₂O atindral charge - + x quillition 2.500 x x Now, ka for above reaction will be ka [N13] [13 ot [Na] 5:55 x 10 = x.a (2:51 reflecting & in denomentos is very small as it

Wo get, 5.55 X10-9x2.5 -2? 3.72 xlong Now Thus [H3O+] = 3.72x10-4 Now, pH = -log [H₂ot) -log [3.72 40 ] ph= 3.42 is required ph 06 Au Value of pka of acetic acid -4.74. 4.88 of acid Value of pka Since P i acidic of and acid a ginen by ph pH = pka + log [bar] [Ard pl & 1 which shows pka. Thus it will be weaker than acetic acid atte sample 4.6x10th concentration of [ht] in blood We know, pH = -log[ht] .pH = -log [4.6x10"] pl = 3.33

At 25°c 1 PH + POH = 14 .pot = 14-ph - 14-3.33 pon- 10.67 Now Pok= -log[or] [04] ph =10 lo- 1067 [04] = 2.13x10 "M Since pH i acidic. of sample to less than 7, it xom

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