Question

Question 1 (0.33 points) pH 4 buffer sodium formate (NaHCoo) ammonia (NH3 sodium acetate (Na[CH3COOD ammonium chloride (NH4)Cl sodium hypochlorite (NaCLO) hypochlorous acid (HCLO) L formic acid (HCOOH) pyridine (C5H5N) pyridinium chloride (C5H5NH)Cl

Can someone help me out. I really need it! Thank you

Answer 1

Ans 1. The acidic buffer solution containing formic acid and sodium formate would lies in the pH range around 4.

Ans 2. The acidic buffer solution containing Hypochlorous acid and Sodium Hypochlorite lies in the pH range around 7.

Ans 3. The basic buffer solution containing Ammonia and Ammonium chloride lies in the pH range around 9.

Ans 4. Buffer Ratio, as explained in the question is ratio of concentration of Conjugate base to conjugate acid. For Formic acid, it is:
Buffer ratio = ( [HCOO^-]/ [HCOOH])

Now the given volume is 0.100 L and moles of formic acid are 0.30,
So, concentration for formic acid is
[HCOOH] = 0.30/0.10 mol/L = 3 moles/L

Given moles of sodium formate is 0.60
Considering the complete dissociation of Sodium Formate, moles for formate ion is also takes as 0.60.
So, the concentration of conjugate base of formic acid, i.e. formate ion is
[HCOO^-]= 0.60/0.10 = 6 moles/L

Buffer ratio, according to the equation mentioned above, for formic acid and sodium formate comes out to be
6/3 = 2.

Ans 5. Acc to the given Henderson–Hasselbalch equation,

${\mathrm {pH}}={\mathrm {p}}K_{{\mathrm {a}}}+\log _{{10}}\left({\frac {[{\mathrm {A}}^{-}]}{[{\mathrm {HA}}]}}\right)$

Where A^- is conjugate base and HA is conjugate acid.

The value for desired pH is given as 3.70 and the pK_​a value for formic acid is given as 3.74,
Thus the equation reduces to
3.70-3.74 = log ( buffer ratio)
=> -0.04 = log (buffer ratio)
=> buffer ratio = 10^(-0.04) = 0.9120

Ans 6. Buffer solution can be acidic and basic in nature. So it can be prepared by using the following methods out of those listed in the question.

1. By partially neutralising the weak acid solution by addition of strong base.
3. By adding appropriate quantities of weak acid and its conjugate base.
5. By partially neutralising a weak base solution by addition of strong acid.

Ans 7. The values given are to be determined experimentally.
The correctly matched value are:
1. Ka for H₂PO₄^-= 6.2 * 10^-8
2. Ka for HPO_​4^2- = 3.6* 10^-13
3. pKa for H₂PO₄^- = 7.21
4. pKa for HPO_​4^2- = 12.44

Ans 8. The molar mass of potassium phosphate(K₃PO₄)
=>3(atomic mass of K) + atomic mass of P + 4(atomic mass of O)
=>3 (39) + 31 + 4(16)
=>212

​