Answer:
4.3
Explanation:
Data provided in the question as per the question is as follows
[tex]k_b = 1.6 \times 10^{-6}[/tex]
Now we use the log in both the sides
So,
[tex]Pk_b = -log\ k_b = -log (1.6 \times 10^{-6})[/tex]
[tex]Pk_b[/tex] = 5.795
And, C = 0.0015
log c = -2.823
Now pH is
[tex]= \frac{1}{2} (Pk_b - log\ c)[/tex]
[tex]= \frac{1}{2} (5.795 - (-2.823))[/tex]
= 4.3
Hence, the pH of the solution of 0.0015 M morphine is 4.3 and the same is to be considered by applying the above formulas and the calculations part
