Determine the ph of a 500.0 ml solution of 353.0mg nacn (ka = 2.1 x 10-9 for hcn).

First, find the concentration of the NaCN solution in terms of molarity. The molar mass of NaCN is 49.0072 g/mol

Molarity = mol solute ÷ L solution

Molarity = (353 mg * (1 g / 1000 mg) * (1 / 49.0072 g/mol)) ÷ (500 mL * (1 L / 1000 mL))

Molarity = 0.0144 M NaCN

NaCN is a salt from a weak acid HCN and a strong base NaOH. The reaction is as follows:

HCN + NaOH --> NaCN + H2O

Upon hydrolysis, it becomes:

+  --> HCN +  

Initial: Amount of is 0.0144 M, is in excess, and the products are 0.

Change: This is unknown. Let this be x. For the , it would be –x, while it would be +x for the products HCN and .

Excess: This would be Initial + Change. Thus for , it would be 0.0144 – x. For both products, it would be x for each.

The hydrolysis constant () would be the amount of products over the reactants. Thus,

= (HCN)*() / () = x2 / (0.0144 –x)  --> equation 1

is also equal to the equilibrium constant of water ( = 1 x 10^-14) divided by .

= 1 x ÷ 2.1 x  --> equation 2

Substituting equation 2 to equation 1:

÷ (0.0144 –x) = 4.762 x

Solving for x,

X = 2.59 x

As mentioned before, x denotes the products HCN and . Thus the amount of is 2.59 x

To find pH,

pH = 14-(-log )

pH = 14 – (-log 2.59 x )

pH = 10.41

Thus, the answer is 10.41.

Q2eqefefgvsdfvfefsvsdvcwef

The problem here is that according to aufbau principle before 3d orbital 4s subshell comes.after filling 3p orbital with electrons you have 4 electrons left, you place two in 4s with opposite spin and the other two in 3d in any of its two orbitals of total five orbitals with the same spin.