Question

1. Determine the pH change when 0.063 mol HBr is added to 1.00 L of a buffer solution that is 0.326 M in HF and 0.211 Min F. pH after addition - pH before addition = pH change = 2. A buffer solution contains 0.469 M CH3NH3Br and 0.323 M CH3NH2 (methylamine). Determine the pH change when 0.091 mol KOH is added to 1.00 L of the buffer. pH after addition - pH before addition = pH change = 1

Answer 1

HE acid - Ht + F conjugate base determin ph of buffer befor addition of HBr can be by using Henderson equation, pH = -log ka & Log [Conjugate base] [cweak and ka for HE is 6.6x10mg. Thus, pH = -log 6.6410 & Log 0.211 0.326 ph= 3.1804 + (-0.18) & ph= 3. 00 04 23. - Iph=3 Now, for calulating ph after addition. As we are adding HBr in the buffer prepared by 0.3 26 mole HF (Since volume is IL and nacxv.) and 0.211 mol F, the concentration of it ions increases as HBr is a strong electrolyte

HF . This the F decreases. Hence, F will combine with Ht to give concentration of HF increases and that of final concentration of these will be [MF] = 0.326 +0.063 - 0.389 F = 0.211 - 0.063 = 0.148 0.063 is taken as 0.063 mol. HBn are added and Imole M Br gives I make it so 0.063 mole Hor gives 0.063 mole of At Now, pH after addition, wing Henderson equation. pH = -log ka + log [conjugate base] [weak a cid Note that now you have to take find concentrations. - pH- - Log 6.6 Xory + log 0.148 0.389 ph = 2.7614

Thus, according to the question ph change - pH after addition - = 2.7614-3 1 ph change. = -0.2386) pH before addition, ② This case is of basic buffer Ch NH2 +h2o weak bas Cha Nnt conjugate ton acid pon for this is given by. =- log ko & pon Log [conjugate a vid] [weak bax] ko for CH₃NH, is 4.4 X long pon - Log 4.4x10 + log 0.469 0.323 & Now pon = 3.35 +0.161 1 por = 3.50] as ph+ pon = 14 pl=14-pon = 14 -3.511 +]PH = 10.489)

Now w as we are adding kon which is a strong. The concentration of on ions increases and nce Cha Nhy will combine with on to give chown. Thus the concentration incentration of of CU₃NH2 will increase and cl₂Nt will the concentration of chynna will decrease As Imole kot gives I mole on so, o.og/mole kon will give 0.091 mole on This finel concentration of Cha Nh, and coz what could be LCH3NH2 = 0.323 + 0.0g) = 0.414 Eng Nng?] = 0.469-0.091 = 0.378. This find pon will be. pon = -log ko & Log [Conjugate aud] Tweak base - pon = - tag 4.4x10ny & Log 8.378 0414 a pon = 3.35-0.039 [pon = 8.310) This, pn= 14-3.311 pH = 10.689

Thus ph before addition ph change = pn after addition - pn change = 10.689 - 10.489 [PH change = 0.2]
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