Question

Can someone please explain to me in detail how to do this problem? Please show all of the work and do not skip any steps!

What is the pH at the equivalence point in the titration of 10.0 mL of 0.35 M unknown acid HZ with 0.200 M NaOH? K_a = 2.4x10^-7 for the unknown acid HZ

A] 7.00

B] 10.1

C] 4.14

D] 9.86

Answer 1

Naz laq) + H2018) I mole of Naot neutralize I mole of HZ Hz (aq) + NaOH(aq) → weak acid strong base At equivalence point Mivi = M2 V ₂ M represent concentration and V represent volume for Hz for NAOH Mi = V = 12 = 0.35M 10.0mL Mix M2 M2 = 0.2009 V2 = ? 0.35m x 10.0mL 0.200M = = 17.5 mL Anthrawaddi ) volume of NQOH : 17.5 mL mmol (milimol) concentration (M) X volume (mL)! mmol of H2 = 0.35M X 10.omla 3.5 mmol mmolo NaOH = 0.200 M X 17.5 mL = 3.5 mmoll mmol Before u HZ laq) + NaOH (aq) Naz (aq) + H2011) 3.5 3.5 change - -3.5 0 +3.5 3.5 After - - 3.5 0 Inmol of Naz final volume Naz] : final volume 1000ml + 17.5 ml [Naz) = 27.5 mL 3.5 mmol = 27. 5ML 0.1273 M Noz i salti democrates completely in aqueous solution Naz laq) → Nat cagl + z-lag) After disocration 0.1273M 0.1273 M [2-] = 0.1273 M ka for Hz = 2.4810-7 Kb for a = kW à 10-14 = 3.70 8 10 8 ka 2-4 X1o-1

OH(aq) - +x х ICE Table concentration (M) z-lag) + H₂0 121 Hz lag) + 0.1273 c -X - + x E 0.1273-4 - x K6= [HZ] [OH-] 3.70 X 10P & scxoc (2-) 0.1273-10 As value of kB is very small, so we can neglect x in comparison to 0.1273 - 3.70 X 10-8 = x² ³x² 0.41 X 10-8 0.1273 4.16 = 6.86 X10 -SM [OH-] = x= 6.86 X 10 -5 M pot = -log(OH-] = -log (6.86X10-5) = БН АБоҢ = Tч bH = 14-bol = 14-4.16 = 9.84 (pH=9.84 This value is near to 9.84 = 1PH = 9.86 Hence, option D) is correct.