Question

4) A 40.0 mL sample of 0.100 M HCl is titrated with 0.100 M NaOH. Calculate the pH after the addition of each of the following volumes of NaOH: (a) 0,0 mL, (b) 20.0 mL, (c) 40,0 mL, (d) 60,0 mL. A plot of the pH of the solution as a function of the volume of added titrant is known as a pH titration curve. Using the available data points, plot the pH titration curve for the above titration,

Answer 1

The acid and base reacts to give salt and water. The data is first collected and the number of moles for each of the acid and the base for different volumes of base is calculated, which gives us the value of proton concentration or hydroxyl concentration. By using the concentration values, we can find out the pH of the solution. Then a plot is drawn, and the whole process is shown in the pictures below:

Answer: Tibrution of Strong Acid vs Strony . Base Het Nach A ad Base H+ + OH Nach +H₂O Salt Water HO Griven data: Hum 0.1M Nao. olm hom na cid=4 [ht] for Strong Aad f Strony Base ht] = . n Acid - neare (40 + V) mL CH+] = u nease notu) And. [pH = = log Cht] - @ For volumes of Nach : (a) v=0.0 mL n Base = 0 = 104 ...: [HP] vsem = 4 = 1 40 10 → bH = r lo q clony) =+1. - > b=1

76 n = 200mL = & nRose = balmx20mL Base = 2 :: [ht] vezom =__4-21 = 21 = 1 140+20) 601 30 pH = -log[ 17 = +1.48 _ pH = 1.48 = +1.48 L 30 1 C) V= 40 • Om L - пеале = 0.1мх Чоunt Nipase = ? - .: [H+) - 4-4 = 0 AL 40440 3 pH = -log (0) = undefined But, For equal amounts of Acid and Base, the pH becomes Neutral. - do HbH = ZL d) N= 60.0mL 7 nBase = 0.1 MX 60ML Base = 6 - COH-] = 6-4 = q = 0.02 60+40 100 - 5 boks clog (0.02) => poh= 1.71

As pH = 14-60Hz 14-17 = 12.3 5bH = 12:31 We of 2. Steep Lusse ng → Methyl orange detects. Equivalence point : - Phenolphthalein_detecta and point t W o 20 no to so = - volume of 0.1 M NaOH cm)