Answer:
Concentration = 0.14 M
Explanation:
Given data:
Volume of base = 19.62 mL
Molarity of base = 0.341 M
Volume of acid = 25.00 mL
Molarity of acids = ?
Solution:
Number of moles of base = Molarity × volume in litter
Number of moles = 0.341 mol/L × 0.02 L
Number of moles = 0.007 mol
Concentration of acid:
Concentration = 1/2 (0.007 mol) / 25 mL × 10⁻³ L/mL
Concentration = 0.0035 mol / 25 × 10⁻³ L
Concentration = 1.4 × 10⁻¹ M
Concentration = 0.14 M
<span>This would be the atomic mass. In an atom of carbon-12, there are 6 protons and 6 neutrons at rest (electrons have a negligible mass and are usually not part of the overall mass calculation). All atomic masses are based off the measurements of this specific iteration of carbon.</span>
NH3(aq) + HCl(aq) ⇒ NH4Cl(aq) >>> (1)
∵ C = n/V; C= concentration, n= No. of moles, and V= volume (L)
∴ n = C*V, n(HCl) = 0.050*(50/1000) = 0.0025 moles
n(NH3) = 0.050*(50/1000) = 0.0025 moles
So, the limiting no. of moles is 0.0025 moles >>> (2)
∵ NH3 is weak base, and HCl is strong Acid (and have the same number of moles) >>> So, without any calculation we can notice that the formed salt (NH4Cl) is acidic salt and the pH is less than 7.
From (1) and (2), The no. of moles of NH4Cl is 0.0025 moles >>> (3)
∴ the concentration of [NH4Cl] = 0.0025 / (total volume per L)
= 0.0025 / ((50 + 50) / 1000) = 0.025 M
NH4+(aq) ⇔ NH3(aq) + H+(aq) >>> (4)
(0.025 - x) (x) (x) >>> (5)
∵ Ka = [NH3] [H+] / [NH4+] >>>> (6)
Ka = Kw / Kb, Kb = 1.8 * 10^-5 >>> (7)
∴ Ka = 10^-14 / 1.8*10^-5 = 5.56*10^-10 >>> (8)
From (4), (5), (7) and (8)
Ka = 5.56*10^-10 = (x * x) / (0.025-x) , we will assume that (0.025 - x) = 0.025
∴ x^2 = (5.56*10^-10)(0.025) = 139*10^-13
∴ x = 3.73*10^-6 = [H+]
∵ pH = - log [H+]
∴ pH = - log 3.73*10^-6 = 5.43
