Astudent intends to titrate a solution of a weak monoprotic acid with a sodium hydroxide solution but reverses the two solutions and places the weak acid solution in the buret. after 23.75 ml of the weak acid solution has been added to 50.0 ml of the 0.100 m naoh solution, the ph of the resulting solution is 10.50. calculate the original concentration of the solution of weak acid

Hey there!

weak acid = HA

is titrated with NaOH

then :

V = 23.25 mL of weak acid

V = 50 mL base, M = 0.1 M base

pH= 10.50

find original solution of acid :

mmol of base initially = MV = 0.1*50 = 5

mmol of acid added = let it be "x"

After addition  :

mmol of base left = 5-x

from pH  :

pOH = 14-pH = 14-10.50 = 3.5

[OH-] = 10^-pOH = 10^-3.5 = 0.0003162 M left

V total = 23.25 + 50 =  73.25mL

mmol of OH- left = MV = 73.25*0.0003162 = 0.023161

Then  :

mmol of acid added = (5-0.023161) = 4.976839

mmol of acid added was = 4.976839

[HA] = mmol of aid/Vadded

= 4.976839 / 23.25 = 0.21405 M

[ HA ] = 0.21405 M

Hope that helps!

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.