raon added at 25% neutralization? Calculate the expected initial pH of your mono

Question

raon added at 25% neutralization? Calculate the expected initial pH of your monoprotic acid solution, using the literature pk, value and the actual concentration of your solution. Hint: what is the concentration of 2 g of your unknown acid in 50 ml. of water? Is this a solution of pure weak acid, pure weak base, or a mixture of the two? Questions for Member B .Calculate the pK, of your monoprotic acid using the pH observed when approximately 75% of the acid originally present has been neutralized. Hint: what is the [base]/[acid] ratio at 75% neutralization? What is the volume of NaOH added at 75% neutralization? Calculate the expected equivalence point pH of your monoprotic acid solution, using the literature pK, value and the actual concentration of your solution. Hint: what is the concentration of 2 g of your unknown acid in 50 ml. of water? Is this a solution of pure weak acid, pure weak base, or a mixture of the hwo?

Explanation / Answer

you used 1.01 M , 0.031 Liters of a strong base , calculate the number of moles of naoh you used

moles = molarity * volume = 1.01 * 0.031 = 0.03131 moles of NaOH

since this is a monoprotic acid then it means that 1 mole of naoh is required to neutralize 1 mole of the weak acid, so there are 0.03131 moles of acid in 2 grams

calculate the molarity with the given information

molarity = 0.03131 / 0.05 = 0.6262 M

since this has a pka then it is a weak acid, lets make an ice chart

....................HA ===== .................H .............+ ..............A

I................0.6262.........................0.................................0

C................-x...............................+x...............................+x

E.............0.6262-x.......................x...................................x

aA + bB cC + dD

Equilibrium constant is the relationship of products and reactants expressed like this

K = [A]a [B]b / ( [C]c[D]d )

K is the equilibrium constant

ka = 10-pka = 0.0001412

Ka = [H] [A] / [HA]

0.0001412 = x2 / (0.6262 - x)

if you solve this equation you will find a value of x = 0.00933 M

since H is equal to x

then ph = - log (0.00933) = 2.03

*only 1 question at a time please...

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