What is the ph of a 0.25 m solution of nacn? (the pka of hcn is 9.21.) (a) 4.91 (b) 8.61 (c) 11.30 (d) 13.40?

Answer is: pH of solution of sodium cyanide is (c) 11.30.

Chemical reaction 1: NaCN(aq) → CN⁻(aq) + Na⁺(aq).

Chemical reaction 2: CN⁻ + H₂O(l) ⇄ HCN(aq) + OH⁻(aq).
c(NaCN) = c(CN⁻) = 0.25 M.

pKa(HCN) = 9.21.

Ka(HCN) = 10∧(-9.21).
Ka(HCN) =  6.16·10⁻¹⁰.

Ka(HCN) · Kb(CN⁻) = 10⁻¹⁴.
Kb(CN⁻) = 10⁻¹⁴ ÷ 6.16·10⁻¹⁰ 

Kb(CN⁻) = 1.62·10⁻⁵.
Kb = [HCN] · [OH⁻] / [CN⁻].
[HCN] · [OH⁻] = x.
[CN⁻] = 0.25 M - x..
1.62·10⁻⁵ = x² / (0.25 M - x).
Solve quadratic equation: x = [OH⁻] = 0.002 M.
pOH = -log(0.002 M) = 2.70.
pH = 14 - 2.70 = 11.30.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

i think it's c. if you look search electronic symbol on google and then click on wiki it shows you a