What would be pH of a 0.01 M solution of
a very expensive weak base, unobtainamine,
whose pKb is 3.8?
1. 2.9
2. 14
3. 2
4. 7
5. 11.1
pKb = 3.8.
Kb = -log 3.8
Kb = approx 1.6E-4 but you should get a better answer than that.
.......BNH2 + HOH --> BNH3^+ + OH^-
I......0.01............0.......0
C......-x..............x.......x
E......0.01-x..........x.......x
Kb = 1.6E-4 = (x)(x)/(0.01-x)
Solve for x = (OH^-) and convert to pH.
Note: YOu may need to use the quadratic formula as 0.01-x may NOT be = 0.01. You will need to try it.
To calculate the pH of a solution containing a weak base, we need to use the pKb value. The pKb is given as 3.8, which is the negative logarithm (base 10) of the equilibrium constant of the base dissociation.
First, let's calculate the Kb value using the pKb value:
Kb = 10^(-pKb) = 10^(-3.8)
Now, let's set up the equilibrium expression for the dissociation of the base, unobtainamine (UNO), as follows:
UNO + H2O ⇌ OH- + UNOH
The Kb expression is:
Kb = [OH-][UNOH] / [UNO]
However, we know that [UNO] is equal to the initial concentration of the base, which is 0.01 M.
Since UNO is a weak base and undergoes partial dissociation, we can assume that [OH-] formed is x and [UNOH] formed is also x.
Therefore, the expression for Kb becomes:
10^(-3.8) = x * x / 0.01
Rearranging the equation, we get:
x^2 = 0.01 * 10^(-3.8)
x^2 = 1.58489319 * 10^(-5)
Taking the square root of both sides gives us:
x ≈ 0.00398
The concentration of hydroxide ions ([OH-]) is approximately 0.00398 M.
To calculate the pH, we need to find the pOH first:
pOH = -log10([OH-]) = -log10(0.00398)
pOH ≈ 2.4
Finally, we can find the pH using the equation:
pH = 14 - pOH = 14 - 2.4
pH ≈ 11.6
So, the correct answer is option 5) 11.6.
To find the pH of a solution, we need to use the expression for the equilibrium constant of the base:
Kb = [OH-][HB+]/[B]
Where:
- "Kb" is the base dissociation constant
- "[OH-]" represents the hydroxide ion concentration
- "[HB+]" represents the concentration of the conjugate acid of the base
- "[B]" represents the concentration of the base itself
Since the pKb value is given as 3.8, we can convert it to Kb using the equation:
pKb = -log10(Kb)
3.8 = -log10(Kb)
By rearranging the equation, we can determine the value of Kb:
Kb = 10^(-3.8)
Now, let's assume that x is the concentration of OH- (which is equal to [HB+]) and 0.01 M - x is the concentration of B. We can substitute these values into the expression for Kb:
10^(-3.8) = x * x / (0.01 - x)
Simplifying the equation, we have:
10^(-3.8) = x^2 / (0.01 - x)
Next, let's solve for x by multiplying both sides of the equation by (0.01 - x):
10^(-3.8)(0.01 - x) = x^2
0.00015849 - 10^(-3.8)x = x^2
Rearranging the equation:
x^2 + 10^(-3.8)x - 0.00015849 = 0
Now, we can solve this quadratic equation to find the value of x, which represents the concentration of OH- and [HB+]. Once we have that, we can calculate the pH.
After solving the quadratic equation, we find that x is approximately 0.01163 M, which is the concentration of OH- and [HB+]. Since the concentration of OH- is the same as the concentration of H3O+ (the hydronium ion) in a basic solution, this concentration can also represent the concentration of H3O+.
To find the pH, we can use the equation:
pH = -log10([H3O+])
pH = -log10(0.01163)
Calculating this, we find that the pH is approximately 1.92.
Since none of the provided answer choices match this value, we can conclude that none of the options are correct.