The answer for this issue is:
The chemical equation is: HBz + H2O <- - > H3O+ + Bz-
Ka = 6.4X10^-5 = [H3O+][Bz-]/[HBz]
Let x = [H3O+] = [Bz-], and [HBz] = 0.5 - x.
Accept that x is little contrasted with 0.5 M. At that point,
Ka = 6.4X10^-5 = x^2/0.5
x = [H3O+] = 5.6X10^-3 M
pH = 2.25
(x is without a doubt little contrasted with 0.5, so the presumption above was OK to make)
I believe dimensional analysis
Answer:
Mass = 76.176 g
Explanation:
Given data:
Mass of lead(II) chloride produced = 62.9 g
Mass of lead(II) nitrate used = ?
Solution:
Chemical equation:
Pb(NO₃)₂ + 2HCl → PbCl₂ + 2HNO₃
Number of moles of lead(II) chloride:
Number of moles = mass/molar mass
Number of moles = 62.9 g/ 278.1 g/mol
Number of moles = 0.23 mol
Now we will compare the moles of lead(II) chloride with Pb(NO₃)₂ from balance chemical equation:
PbCl₂ : Pb(NO₃)₂
1 : 1
0.23 : 0.23
Mass of Pb(NO₃)₂:
Mass = number of moles × molar mass
Mass = 0.23 mol × 331.2 g/mol
Mass = 76.176 g