Question

For the titration of 25.0mL of 0.20M hydrofluoric acid with 0.20M NaOH, determine the volume of base added when pH is (a)2.85 (b)3.15 (c)11.89?

Answer 1

can i tell you for the first part only?

and you can solve the same way for the next parts?

Answer 2

tell me how i can approach all 3

Answer 3

ok.. dude! wait!

Answer 4

first part : pH = 2.85
i.e. -log(H+) = 2.85.
so, [H+] = 1.4*10^-3
this is the resultant [H+] ion concentration.. by mixing 0.025L of HF and say V litres of NaOH.

so. the resultant number of moles of H+ = concentration * volume
= 1.4*10^-3 *(0.025 +V)

and, HF dissociates as:-

\[HF \leftarrow \rightarrow H ^{+}+F ^{-}\]
0.2 0 0 ...............(initially)
0.2-X x x ..................(at time t)

and we know, \[K _{a}(HF) = \frac{ x ^{2} }{ 0.2-x }\]
Ka(HF) = 3.47 *10^-4

solving this for x, we get x = 0.00816.
thus, [H+] = x = 8.16*10^-3

NUMBER OF MOLES OF H+ FROM ACID = CONC. * VOLUME
8.16*10^-3 * 0.025 = 2.04*10^-4

number of moles of OH- = conc. of base * volume taken = 0.2*V

as this no. of moles of OH- react with H+ , we will subtract this 0.2V from 2.04*10^-4 and put it equal to 1.4*10^-3(0.025+V)

solving this for V, you will get the answer.

i.e. you need to solve the equation.

2.04*10^-4 - 0.2V = 1.4*10^-3 ( 0.025 + V)
the ans. will come out to be 8.3 *10^-4 Litres.. i think

Answer 5

see if this helps.

Answer 6

for first part ans = 0.837 mL
for seconddd part ans == 0.928 mL
third padrt 1.02 ml

are these the answers?????????????????

Answer 7

are these the answers?