A 500.0 ml buffer solution is 0.10 m benzoic acid and 0.10 m sodium benzoate has an initial ph of 4.19. what is the ph of the buffersolution upon addition of 0.010 mol ofnaoh? the kafor benzoic acid is 6.5 • 10-5

Question

jalenthomas9442 jalenthomas9442

17-11-2017
Chemistry

contestada

A 500.0 ml buffer solution is 0.10 m benzoic acid and 0.10 m sodium benzoate has an initial ph of 4.19. what is the ph of the buffersolution upon addition of 0.010 mol ofnaoh? the kafor benzoic acid is 6.5 • 10-5

Respuesta :

Otras preguntas

Simplify: 2m – 6 ≤ 18. Group of answer choices m ≥ 6 m ≤ 12 m ≤ 6 m ≥ 12

After the fall of Rome the Muslim wanted to destroy everything romanTrue or false

Which of the following is a measure of the total kinetic and potential energy of the particles in a substance? A. Latent heat B. Specific heat capacity C. Tempe

Spanish home work please hep

y=4x+6 2y=8x+12 what’s the solution is it one solution, no solution, or infinitely many solutions

Which of these is not an example of a secondary source.A) diary entry B)newspaper articles .C)text book D) encyclpodia

what was the purpose of the Harlem Renaissance?

Plz show work so I don’t fail this

name the type of transpiration in tobacco,rose,edible pea,rice

True or false? the plebeians in ancient Rome created a written law code

meerkat18 meerkat18 · Answer 1 · 2017-11-30T08:52:39+01:00

new pH = 4.27

Adding 0.01 mol of NaOH would increase the amount (in mol, 0.10) of sodium benzoate and decrease the amount (in mol, 0.10) of benzoic acid. This is because NaOH would react with benzoic acid to form more sodium benzoate. NaOH would get used up in the solution. Adding (sodium benzoate + NaOH) and subtracting (benzoic acid + NaOH) would give you 0.11 mol of sodium benzoate and 0.09 mol of benzoic acid. New pH = pKa + log [mol of sodium benzoate/mol of benzoic acid]. Hope this helps!