what is the pH if you mix 250ml of 3M HCl with 500ml of 1.5M CN-?
i know that:
HCl+CN- ---> HCN+Cl-
250ml converts to .75 moles and 500ml also converts to .75 moles when using the 3M and the 1.5M.
Ka=X^2/m-x
x/1000=.015/750
.02 CN- left.
2.5*10^5=x^2/(.02-x)
2.5*10^5(x^2/(.02-x))
0=x^2+(2.5*10^-5x)-(.5*10^-6)
x=.695*10^-3 [OH-]
4.0*10^-10=x^2/1-x
.75/750=x/1000
0=x^2+94.0*10^-10x)-(4.0*10^-10)
x=1.99*10^-5 [H3O+]
(.695*10^-3)-(1.99*10^-5)
1*10^-14=((6.751*10^-4)[OH-])/([H3O+](6.751*10^-4))
from there i'm stuck.
To determine the pH of the solution after mixing 250 ml of 3M HCl with 500 ml of 1.5M CN-, we can use the following steps:
1. Calculate the total moles of HCl and CN-:
- HCl: 250 ml * 3 mole/liter = 0.75 moles
- CN-: 500 ml * 1.5 mole/liter = 0.75 moles
2. Write the balanced chemical equation for the reaction:
HCl + CN- → HCN + Cl-
3. Calculate the remaining concentration of CN- after the reaction using the mole ratios:
Since 0.75 moles of HCl reacts with 0.75 moles of CN-, there will be 0.75 moles of CN- left.
4. Use the given equilibrium constant expression, Ka = [HCN][Cl-]/[HCl][CN-], to set up an expression for x:
We can assume that the amount of HCN formed is "x" moles, and therefore, the amount of CN- remaining will be (0.75 - x) moles.
5. Substitute the concentrations into the equilibrium constant expression and solve for x:
Ka = (x)(0.75 - x) / (0.75)(0.75)
Solve the quadratic equation to find the value of x. In this case, x = 0.695 * 10^-3 M [OH-].
6. Calculate the concentration of [H3O+]:
Since the reaction is in water, we use the equation Kw = [H3O+][OH-]:
Kw = (1.0 * 10^-14 M^2) = (0.695 * 10^-3 M)([H3O+])
Solve for [H3O+]: [H3O+] = 1.99 * 10^-5 M.
7. Calculate the pH:
pH = -log[H3O+]
= -log(1.99 * 10^-5)
= 4.70
So, the pH of the solution after mixing 250 ml of 3M HCl with 500 ml of 1.5M CN- is approximately 4.70.