Question

5. Calculate the pH of buffer formed by mixing 85 mL of 0.12 M lactic acid with 95 mL of 0.16 M sodium lactate (K, for lactic acid is 1.4 x 10-) a. 3.77 b) 4.03 ) 3.56 d) 4.23 e) 3.66 6. 18 2.5mL of 0.30M AgNO3 is mixed with 7.5mL of 0.015M Na2SO4, should a precipitate of Ag2SO4 form? (Ksp - 1.2x10-5 a) Q - Ksp the system is at equilibrium and the solution is saturated b) Q<Ksp, more solid can dissolve, so no precipitate forms. c) Q> Ksp a precipitate will form. 7. A 35.0-mL sample of 0.150 M acetic acid CH3COOH (K-1.8x 10-) is titrated with 0.150 M MacBook Air # $ % &

Answer 1

kindly post different questions separately.

NO C3Hs 0.3 Lactic acid = C3H603 Sodium Lactate = milimoles (mmol) = concentration (M) x volume (mL) mmol of C3H6O3 = 0.12 M X 85 mL = 10.2 mmol mmol of NaC3H503 = 0.16 M X 95 ml = 15.2 mmol After adding both solution final volume of buffer solution : 85ml + 95m1 = 180 mL [C3H603) = mmol = 10.2 mmol = 0.0567M tinal volume (ml) 180m2 [NOC3 HSO3] = mmol 15.2 mmol = 0.0844 M final volume (ML) 180ml ka - 1.4 X10-4 bko: - 109 kq = -luq (1.4 X10-4 ) = 3.86 Lactic and is weak and so do not cissociate completely in aqueous solution while sodium lactate, salt, climoclate completely in ag wow solution NO C 3 HS03 lag) Nat lag) + (3H5O3- (og) After diesociation 0.0844 M 0.0844M → [C3H503-] = 0.0844 M Using Hendenon Hasalbalch equation. - pH = pka + lug [ Cats 03-] (C3H603] ÞH = 3.86 t log 00844M 0.0567M pH = 4.03 Hence, option b) is correct.