Question

A 41.0-mL sample of 0.217 M CH, COOH solution is titrated with 0.176 M N OH. Calculate the pH of the solution (This problem rHF HNO CH,COOH HOCI HOBI HOCN HCN H2SO4 K = 72 x 10- K = 45 x 10 K = 1.8x10 K = 3.5x10 K -2.5 x 10 K= 3.5 x 10 K, 4.0x 10-10

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@ ch coolH+ 40 ~ CH coo thot 0.212 1 0 0 I et cootika Kaal8x105 X 1 x EL 0.217-8 1-1 [ch coo] [Hot ka² ECH coolf 0.217-X = 0.③ nach: Molarity = 0.176M volume = 20-3 m) = 0.02032 no-ot moles = 0.1764 0-020320.0035728 eff, cooltt Nach > CH₃ Coona + H₂O(2) Nadh molarity = 0.126M volume & 23.2m120.02322 no ot moles = 0.126 40.0232= 0.00 40832 et coolt & NaOH > CHCOONa tho 0.00. Naolt molarity = 0.126M volume = 27-7 ml = 0.022772 no ot moles = 0.176 4 000 277= 0+0048752 CH COOH + NaOH CH COONa theo IÔ NaOH Molarity 20-126M volume = 50-6m1= 0.0506L noot moles=0-126x8.0506 = 0.0089856 chcoolt + NaOH - CH₂ Coonat to Il 0.0088(9) Naoff molarity 2 0.126 volume = 52.3 mia 0-05232 notot moles = 0.126x0-0526= 0.00 92048 CH cooft WooH CH coonat to Il 0.0

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