How many g. of NaCN would you need to dissolve in enough water to make exactly 375 mL of solution w/ a pH 10?

Question

Hey put this up earlier and never got a response i really need someones help

Answer 1

HCN is a weak acid with pKa=9.2
Thus CN- will hydrolyze according to the reaction

.. .. .. .. .. .. .. CN- +H2O <=> HCN +OH-
initial .. .. .. .. .C
React .. .. .. .. x
Produce .. .. .. .. .. .. .. .. .. .. .. .. x .. .. x
At Equil. .. .. C-x .. .. .. .. .. .. .. .. x .. .. x

Kb= [HCN][OH-]/[CN-] = x^2/(C-x) => C=(x^2+Kbx)/Kb (1)

but Kb=Kw/Ka= Kw/(10^-pKa)= 10^-14/10^-9.2 =10^-4.8
and since pH=10 => pOH=14-10=4 => [OH-]=10^-4
and [OH-]=x
Taking all these into account substitute in (1)

C=((10^-4)^2 + (10^-4.8)*(10^-4)) / (10^-4.8) =7.31*10^-4 M

The MW of NaCN is 49
C=mole/V
mole=mass/MW so C=mass/(MW*V) =>
mass=C*V*MW = (7.31*10^-4)*0.375*49 = 0.0134 g

Please check for numerical errors

Answer 2

you are able to locate the atomic sort i think (i dont understand the kind on the backside of the sq. in the periodic table) like hydrogen (H) is a million and get Na,C,N and upload them up then im not precisely particular what you do after that enable me verify in my notes i in simple terms have been given accomplished doing those