thank soooo much!!! e Truncatec Outine-Essay Two, x a MasteringBiology: User Set

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thank soooo much!!!

e Truncatec Outine-Essay Two, x a MasteringBiology: User Settin × Cheap Flights, Anne T ckets x ( G meiosis ll s.edu/courses/57153/quizzes/252188/take Question 15 5 pts Calculate pH after the addition of 0.10 mole of NaOH to a 500 mL buffer made up of 0.500 M HF (Ka-6.8 x 104) & 0.500 M NaF? Assume no change in volume of solution upon addition of base. 02.77 ? 2.47 O 280 DQuestion 16 5 pts Calculate pht after the addition of 100.0 ml of 1.0M HCl to a 500 mL buffer made up of O 500 M HF (Ka-6 8x ¡0*) & 0500 M NaF? 3.54 ?2.80 3.57 88B9888

Explanation / Answer

Calculation of pH of buffer solution :-

Given :-NaOH = 0.10 mole, HF =0.500 M ; NaF = 0.500 M,    V=500ml or 0.5 L ; Ka = 6.8 (10-4)

After addition of 0.10 mole of NaOH, there is reaction between NaOH and HF.HF is not a strong acid,it is weak acid.

HF(aq) + -OH(aq)     <=====>   H2O(l)     +   F-(aq)

New initial concentration are:- HF =0.400 M ; NaF = 0.600 M

HF(aq) + H2O(l)     <=====>   H3O+(aq)     +   F-(aq)

Ka = [H3O+][F-] / [HF]

[H3O+] = Ka [HF]/[F-]

[H3O+] = 6.8 (10-4) [0.400] / 0.600

[H3O+] = 4.53 (10-4)

pH = - log [H3O+] = - log [4.53 (10-4)]

pH = -[ log 4.53 + log (10-4)]

    = -[ 0.656 +(-4)1]

    = - 0.656 + 4

   pH = 3.344

HF -OH H2O   F- I 0.500 0.10 0 0.500 C -0.10 -0.10 0 + 0.10 E 0.400 0 0 0.600

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