(34) Apply Handersson's equation to get the desired answer-
pH = PKa + log [salt] / [ Acid ]
= -log Ka + log [ Salt ] / [ Acid ]
4.35 = - log ( 1.8 x !0-5 ) + log { [Salt ] / [ Acid ] }
= 4.7447 + log{ [Salt ] / [ Acid ] }
So, log { [Salt ] / [ Acid ] } = - 4.7447 + 4.35
= - 0.3947
& [Salt ] / [ Acid ] = antilog ( -0.3947 )
= 0.4069
hence, the correct answer is -----------------> " b''
--------------------------------------------------------------------------------------------------------------------------------------
(35 ) Find out the volume of HCl solution used by appplying relation-
0.23 x V = 0.17 x 27
V = ( 0.17 x 27) / 0.23
= 19.95 ml
So volume of 0.23 M NaOH used for titration = 19.95 ml which is also equivalent to 19.95 ml. of 0.23M HCl soln.
Volume of HCl left = 43- 19.95 = 23.05 ml of 0.23M HCl soln
but now this volume is present in (43 + 27 ) ie. 70 ml of soln. , hence calculate the molarity of this soln. & find out pH
70 x M = 23.05 x 0.23
= 0.07573M
& pH = - log ( 0.07573 )
= 1.12
hence, the correct answeer is--------------- ( c )
------------------------------------------------------------------------------------------------------------------------------- Pl. post the other question separately, for answering, Glad to help.
However , here is the answer-
(36) Study the equilibria & find out [OH- ] as below-
CH3 COO^- (aq) + H2 O (l) <---------------> CH3COOH (aq) + OH^-(aq)
initial conc. (moles) -------------- 0.09-----------------------------------------------o ----------------0
change ------------------------------ -x ------- -x ----------------------------------- +x ---------------+x
at equilibria ---------------------------(0.09- x ) H2O --------------------------------x ----------------- x
K = { [CH3 COOH] [ OH^- ] } / [ CH3COO^- ]
= (x) (x) / (0.09 - x )
= x^2 / 0.09 (x is usually so small & negligible, hence is ignored frrom the denominator)
Again , K = x^2 / 0.09 = Kw / Ka = 10-14 / 1.8 x 10-5 *
So, x^2 = 49.95 x 10^-12
x = 7.07 x 10^-6
This represents the [ OH- ] also ie. [OH- ] = 7.07 x 10-6
Again [ H+] [ OH- ] = 10 -14
therefore, [H+ ] = 10-14 / 7.706 x 10-6 = 1,4 x 10-9
& pH = - log[H+] = - log(1.4 x 10-9 )
= 8.85
Thus the correct answer is ------------------------ (d)
Note-
* The value of Ka is calculated as 1.8 x 10-5 using the relation-
pKa + pKb = 14
pKb is given , we can calculate pKa
Known to us pKa value = - log Ka
Suppose that you wanted to prepare a acetate ion/acetic acid buffer solution with a pH of...
You need to prepare an acetate buffer of pH 6.29 from a 0.704 M acetic acid solution and a 2.48 M KOH solution. If you have 575 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 6.29? The pKa of acetic acid is 4.76. You need to prepare an acetate buffer of pH 6.29 from a 0.704 M acetic acid solution and a 2.48 M...
Suppose you want to make an acetic acid/acetate buffer to a pH of 5.00 using 10.0 mL of 1.00 M acetic acid solution. How many milliliters of 1.00 M sodium acetate solution would you need to add? The pKa for acetate buffer is 4.75.
pH of acetic acid/acetate buffer pH of pure water Pure solution (no additions) Addition of 10 mL of 0.01 M HCI Addition of 10 mL of 0.01 M NaOH Comment on the changes in pH in the acetate buffer as opposed to that of pure water. What does this tell you about the role of a buffer?
=Assume you have prepared 100.0 mL of a buffer solution using 0.400 mol of acetic acid (pKa = 4.74) and 0.400 mol of sodium acetate. The pH of this buffer solution is initially 4.74. After preparing this buffer solution, you added 55.0 mL of a 1.10 M NaOH solution to your buffer to see what would happen. What will the pH of this new solution be?
A 1.00 L buffer solution with pH = 4.74 is composed of 0.30 mol acetic acid and 0.30 mol sodium acetate. A) Determine the pKa of acetic acid B) If 0.030 mol of NaOH is added, determine the pH of the solution
You need to prepare an acetate buffer of pH 6.25 from a 0.896 M acetic acid solution and a 2.46 M KOH solution. If you have 680 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 6.25? The pK, of acetic acid is 4.76. ml
You need to prepare an acetate buffer of pH 5.21 from a 0.635 M acetic acid solution and a 2.86 M KOH solution. If you have 830 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.21? The p?a of acetic acid is 4.76.
You need to prepare an acetate buffer of pH 5.77 from a 0.671 M acetic acid solution and a 2.12 M KOH solution. If you have 625 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.77? The pK_a of acetic acid is 4.76.
You need to prepare an acetate buffer of pH 6.31 from a 0.848 M acetic acid solution and a 2.32 M KOH solution. If you have 675 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 6.31? The pKa of acetic acid is 4.76.
You need to prepare an acetate buffer of pH 5.37 from a 0.697 M acetic acid solution and a 2.38 M KOH solution. If you have 980 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.37 ? The p?a of acetic acid is 4.76.