Question

A.)Using a 0.25 M phosphate buffer with a pH of 7.4, you add 0.70 mL of 0.53 M HCl to 52 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)

B.)Using a 0.25 M phosphate buffer with a pH of 7.4, you add 0.70 mL of 0.53 M NaOH to 52 mL of the buffer. What is the new pH of the solution? (Enter your answer to three significant figures.)

Answer 1

the pka volver of the Poly are pka, 2016, pkan tol, prann 12.30 z Po as tapo 4.007- PK (B r ashere, pH. PRAV PH = pkart lag The value of Phil z. po- in close to oka value of pkast. vol. So here in pharh halt bescher at pers. weak and deply to consugate bore 4001 present. (a) & According to foroblem. Chapoy & [ HP042] = 0.25...6 Again from tenderson Hanelbach equation we get ashere, 7 cor conjugat CAT L Gas 7.4 pkan & Rog (CB) . Ao wreak aud +) 74 721 dolog tend logtoo 74 - F.ll= 0 TA → CCM - 10019 = USS CA i cao = 1.55 CAI . (APO ) 1955 x [Haply of 0.25- DHL, PO-2 S SC44 pm 7 Cuma 0.25 2.55 typow] Ch Pour & ans/ 2.550°096. . C'HPO4) = (0.15-0.098) 20152

ho xe & swt sei[#002) (A): In 52 o of buffer typ04-Present 0.898 in xoorxx 52 [:(tor) - 0*096M2 5.096x203 mol similarly in some of buffer HP 042- prosent 0152 mol - - Toooww 027 = 7.904 x70 -s mol wow, in dozom of 0.530 He the number of me of Hel - 0.53 mo ng Ozone – 3.71410 wol = 0.371 003 mol Har aill react with #poy- & aill convert to uz poy at 1:1 ratio an 4 & 4 Poun pou & in .. wet poy fresar a flor adding 0.371 4103 was the = (5.09 6 0 371) X163 mod. = 5.467 418 mo & wet & Poy 2 auill be = 17.904-0.371 )xron? mo! = 7.533x10 mo Now, Total volume of solution = (52 of 0.70m & 52.7 me . after adding tel: E the pool 5.467 410 ? mol & [po, 2] 3 52.70M 73330 WOW

6674/mol .: C#po, a-) CAPOV) -71519x19mmer .: - 0726 Voing vendes son Hanello alch emation we get . Се ру: pr. pkan clog Import =) pH = 7:21 + log (0.746) [:pkan - pH = 7.21 - 014 .: PH=707 .: /PH: 7:07 ] (0) Similarly here, szom of 8.530 wart comain NooH = 0.371810 ? mal wart Initial number of moles of H2O - 5.096410-3 molf molls of 4002- = 7.984 x 103 mul wood react auth 4.pour of arill convert to APO 2 at 1:1 ratio. So, na 0.3718103 mol woolt comerta 0.2718103 mol Hapoy- to s.r71810 mol #poy2- . final Hapoy = (5.096 - 0.371 3x10 m3 & 4.725x10-3 mol final HP02 (7.9044 0.371) 428-3 = 8. 275x103 mol

. Dor the buffer splên mole ratio. à eauivalent to molar ratio. in CHPO4 22) 8.275x10 TKPOED 725 Xlores = 1075 Again ascung fienderson, stane balch equation are get pH = pkagt lo & CHPO 42 ] Throul & 721 & log(1975) - 7.21 & 0.24 = 7.45 .: PH = 7: s]