Question

a. A 0.25 mol sample of HBr is added to 2L buffer solution consisting of 0.68HCN NaCN. which species exist after the chemical reactions?
H+
HCN
Na+
CN-
Br-
b. what is the ph of the resulting solution after a 0.25 mol sample of HBr is added to a 2 L buffer solution containing .68M HCN, Ka=6.2×10^-10 & 0.35 NaCN?

Answer 1

Ans :-

(a). Given ,No. of moles of HBr added = 0.25 mol

No. of moles of NaCN = Molarity x Volume in L

= 0.35 M x 2 L

= 0.70 mol

No. of moles of NaCN = Molarity x Volume in L

= 0.68 M x 2 L

= 1.36 mol

Reaction and ICF table of HBr and NaCN is :

..........................NaCN (aq).................+..................HBr (aq) -----------------> HCN (aq) ...............+......................NaBr (aq)

Initial....................0.70 mol........................................0.25 mol.........................1.36 mol..........................................0.0 mol

Change...............-0.25 mol......................................-0.25 mol.........................+0.25 mol.......................................0.25 mol

Final.....................0.45 mol.........................................0.0 mol............................1.61 mol........................................0.25 mol

After the chemical reaction, species which exist are : HCN, Na+, CN- and Br-

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(b). Using Henderson-Hasselbalch equation :

pH = pKa + log [NaCN] / [HCN]

pH = - log Ka + log 0.45/0.25

pH = - log 6.2 x 10-10 + 0.255

pH = 9.21 + 0.255

pH = 9.46

Therefore, pH of the solution = 9.46