Question

PRE-LAB for pH titration of a Strong Acid with Base This is due before the lab begins, Name 1. Calculate the pH of the following solutions: (a) 1 M NaCl Does not dissociate in water to produce either hydrogen or nyot thus it is a neutral Sall, so 7. (b) 1 M HOAc (Ka - 1.8 x 10-5) duce either hydrogen or hyd xde ion, (c) 1 M NHOH (Kb = 1.8 x 10-5) (d) 0.1 M NaOAC (e) 0.1 M NHCI. 2. The pH of a 0.100 M solution of an acid, HA, is 1.70. Calculate the K? (use ICE) 2.Show your work with units and correct significant figures for all questions that involve a calculation 1. Predict the expected pH (<7, >7.-7) at the endpoint(s) of each of the three titrations. Explain your reasoning. a. Titration of a Strong Acid, HCl, with a Strong Base, NaOH Predicted pH (circle one): (a.) 7 (b.) 7(c.) >7 Circle your answer and explain below. b. Titration of a Weak Acid, KHP, with a Strong Base, NaOH Predicted pH (circle one): (a.) <7 (b.) 7 (c.) >7 Circle your answer and explain below.

Answer 1

ph is the negative log of concentration of ittion in a solution. i pHs-log [H] a) 01" Nau : Nall is a salt and produced from NaOH (strong base) and HU (strong acid). So, it is neutral in nature. pH will be 7 b) 1 M ACOH... It is a weak acid, for it ын - , p<a - 4, 193 с , яка - - Jone en Ź log (1.8x10²s) - ² log (1) 2- (-4.744) - 2:372

AM NHAOH :- It is a weak base. for it pon = Ź pro ² loge - = - 2 log ko - ž logic = -1/3 log (18xios) - log (1) = - (-4.744) = 2.372 - ĐH - 11- bº" = (14- 2- 31 2) = 11.628 d) Oil M Na OAC :- It is a salt. produced from NaOH (strong base) AC OH (weak acid) for it, pl = 7+ 1/2 pka t / loge. = 7+ ² (-log ka) it / log C =7-Ź log (18 X10's) + / log (0.1) = 7- (-4.744) + Ź (-1)

= 7+ 2.372 -0.5 = 8'872 e) O.IM NHAU :- It is a salt. produced from strong Aeid tu and weak base nitaon forit, pH = .7 Ź pkb - £ loge : . = 7 - Ź (-log ko) - £ loge :.= 7th log (18x10's) - leglo.) =7+ 5 (-4.744). - (-1) dalah 3 7. - 2:372 +0.5 = 5:128