Naoac hoac

Q1:

A] if you are given sodium salt of acetate as ( H3COONa.3H2O, then

1 Molarity = 1 Moles/ 1 litre = 136.09g/ 1 litre [ since 1 mole = 136.09g for CH3COONa.3H2O]

But if we want to make 0.2M Solution of CH3COONa.3H2O, then

0.2M = 0.2*136.09g/L = 27.22g/L or 2.722g/100ml Thus 2.722g of this salt is needed in 100 ml H2O to make its 0.2 solution.

If you are given acetic acid of = 17.47 Molar (liquid), then using molarity equation, M1V1=M2V2

0.2*litre = 17.47*v2

V2 = 0.0114 Litre = 11.4ml of CH3COOH needed in 1 litre H2Oon 1.14 ml of CH3COOH needed in 100ml H2O for making 0.2M solution.

Now, for making 0.1 Molar buffer solution, add 100ml of each of sodium salt of acetate and acetic acid in beaker. Thus we get 0.1M buffer solution of acetic acid and its sodium salt.

B] If now we add a small amount of HCl to above prepared buffer, then sodium salt of acetic acid starts working as

HCl + CH3COONa CH3COOH + NaCl (Weak acid less contribution for H+, formation)

So acid will increase, but it will be very small due to weak dissociation of CH3COOH.

C] If we add small NaOH to buffer solution, then CH3COOH prepared in buffer solution starts working, and forming weak base as

NaOH + CH3COOH CH3COO + Na + H2O (weak base)

So, basicity increases very-small. i.e. reaction is not formable to increase significant basicity.

Q2:

We will prepare a buffer that 1:1 molar ratio of HOAc:NaOAc

[0.1M HOAc + 0.1M NaOAc] ------- combination is a buffer.

HOAc is acetic acid, is a liquid weak acid. Our target concentration is 0.1 M (M stands for molarity = moles / liter)

acetic acid molar mass = 60.05 g/mol

Mass of acetic acid required to prepare 100 mL of 0.1M concentration = [molarity x molar mass of acetic acid] x [volume in mL /1000]

Mass of acetic acid required to prepare 100 mL of 0.1M concentration = [0.1M x 60.05 g/mol] x [100 mL/1000]

Mass of acetic acid required to prepare 100 mL of 0.1M concentration = 0.6005 g, but acetic acid exists as liquid, hence we need to convert mass in grams to volume in mL considering its density.

acetic acid density = 1.049 g/mL, density = mass / volume

volume = mass / density = 0.6005 g / 1.049 g/mL = 0.57 mL

volume of acetic acid has to be taken = 0.57 mL

{Sodium acetate) NaOAc molar mass = 82.0343 g/mol

Mass of sodium acetate required to prepare 100 mL of 0.1M concentration = [0.1M x 82.0343 g/mol] x [100 mL/1000]

Mass of sodium acetate required to prepare 100 mL of 0.1M concentration = 0.82 g

Procedure:

1. Take clean and dry 100 mL graduated volumetric flask with stopper

2. add 0.57 mL of acetic acid to above flask by using graduated 1 mL glass syringe.

3. Now also add 0.82 g of NaOAc to above flask.

4. add pure distilled water to above flask upto the 100 mL mark.

5. Put the glass lid. Now shake well and gently the mixture to get clear homogeneous solution.

6. HOAc-NaOAc buffer solution is prepared

Q3:

If you have equal amount of weak acid and its conjugate base in a buffer, the pH is always = pKa where Ka is the dissociation constant of weak acid. Here Ka = 1.8X10^-5.

So pKa = -log(Ka) = 4.745.

We would use the Henderson-Hasselbalch Equation: pH = pKa + log (base/acid) for acetic acid, let us assume 1.00 M in acetic acid and 1.00 M in sodium acetate pH = 4.745 + log (1.00/1.00) pH = 4.745 + zero pH = 4.745 The same result would happen no matter what the two concentrations were.

Q4:

Given; ka = [H+][oAC-]/[HOAc]

= [H+] = ka[HOAc]/[OAc-]

= ph = Pka + log [(b/vi)/(a/vi)]

= pih = pka + log [b/a] formula for ph of buffer (1)

now X moles of acid is added. So, moles of acid = (a+x) finally

Phb = pka + log (b/(a+x)) (2)

ph = 2-1 = log (b/(a+x)) – log (b/a)

-0.1 = log (a/a+x)

Hp= 2-1 will be negative because we are adding acid so, pH will decrease

X = 0.259a 1M*Vtotal = 0.259

Xmole = 1M * (100ml + V)

A= 1M*100 = 100m mol

X = 0.259

(100+V)ml = 0.259 * 100m mol

V = 59ml