What is the pH of 10-9 M HCl?
Here , [H+] for acid is < 10 -7 { which is not possible }
so,
total [H+] = 10 -7 + 10 -9
= 1.01 x 10 -7
Thus, pH = - log [H+]
= 6.99
If the volume of 25 cm3 of 0.05 M Ba(NO3)2 are mixed with 25 cm3 of 0.02 M NaF. Will any BaF2 precipitate? [Ksp of BaF2 = 1.7 x 10-6 at 298K]
Number of moles of NaF in 25.0 cm^3
0.02 /1000 * 25.0 = 5.0 x 10^-4
The total volume of the solution on mixing Ba(NO3)2 and NaF solutions
= 25.0 + 25.0
= 50.0cm^3
This means that 1.25 × 10^-3 moles of Ba(NO3) and 5.0 × 10^-4 moles of NaF are now present in 50cm3
.: The concentration of Ba(NO3)2 in solution
(1.25 × 10^-3 )/50 x 1000
= 0.025 mol L^-1
and, the concentration of NaF in solutions
= (5.0x 10^-4)/50x 1000
= 0.01 mol L^-1
Hence, | Ba?+| = | Ba(NO3)2] in solution
= 0.025 x (0.01)^2
= 2.5 x 10^-6
The value of Ksp for BaF2 is 1.7 × 10^-6 Since the value of ionic product is greater than that of the solubility product, precipitation of Ba F, will occur.