Question

An analytical Chemist is titrating 168.9 mL of a 0.6800 M solution of hydrazoic acid (HN) with a 0.6300 M solution of NaOH. The pk of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 66.95 mL of the NaOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added. Round your answer to 2 decimal places. PH -

Answer 1

As this is a buffer . This is solved by Handerson-Hasselbalch equation.

HN3 NaoH NaN3 O Hene a wean, auid sall (NaNa) i produced an lso some ot wmrealcd HN, will be preseul". We will use Handerson- Hassel bal ch equathiau a uis 1s a buffer Given, 69 mL 06S0OM tydrazwic aid 66.95 ml of D 6300 M NaoH Moles ot tydazoic aid 0.680MX 168.4 mL O-14 moles Moles t NaoH O 630 M X 66.45 L = 0 042 Moke (Ak A M Mols ,Henue mo k = MXL Hene th reauhin will produe o042moles NaN nd (o-114-0042) mali of HNs renaininy 072 moler HNremaln Hene

Now tohl Voluune beomes(169.4 +66.45)m 835 85 ml - D.Q8585 L O.235 L Coaludlabing waolasily o ea cowponeu, Now HN3 0 305 M a35.85 m NaNg 1 935.85 m Now Buosh aing -Haselblh equehän these Value in Hawclesso sat CAcad pt pka +loy pka =412 NaN] (o =4121 1 CHN21 pH 4.2 + Co 305 pH4.12-o 23 pH4-49