Solution A:50cm^3 of 1.0mol dm^-3 hydrochloric acid,HCl
Solution B:10cm^3 of 2.0mol dm^-3 hydrochloric acid,HCl
(a)Calculate the number of moles of hydrochloric acid,HCL in each solution.
(b)Solution A and B are mixed together.
(i)Calculate the number of moles of hydrochloric acid,HCL in the final solution.
(ii)What is the volume of 2.0mol dm^-3 ammonia,NH3 solution needed to neutralise the final acid solution?
A. mols soln A = mol/dm3 x dm3 =?
mols soln B = same thing.
Bi. total mols HCl = mols of A + mols B
Bii.
NH3 + HCl ==> NH4Cl
mols HCl you have.
From the coefficients in the balanced equation, you know mols NH3 = mols HCl
Then mols NH3 = mol/dm3 x dm3. You know mols and mol/dm3 NH3, solve for dm3 NH3.
To solve this problem, we'll use the concept of moles and the stoichiometry of the reaction between hydrochloric acid (HCl) and ammonia (NH3). Let's break it down step by step:
(a) Calculate the number of moles of HCl in each solution:
In Solution A:
Volume (V) = 50 cm^3 = 50/1000 dm^3 (convert cm^3 to dm^3)
Molarity (M) = 1.0 mol dm^-3
Number of moles (n) = V * M
n(A) = (50/1000) * 1.0 = 0.05 moles of HCl
In Solution B:
Volume (V) = 10 cm^3 = 10/1000 dm^3 (convert cm^3 to dm^3)
Molarity (M) = 2.0 mol dm^-3
Number of moles (n) = V * M
n(B) = (10/1000) * 2.0 = 0.02 moles of HCl
(b) After mixing Solution A and Solution B together, let's find the number of moles of HCl in the final solution:
Number of moles in final solution = n(A) + n(B)
n(final) = 0.05 + 0.02 = 0.07 moles of HCl
(i) To neutralize the final acid solution, we need to use the stoichiometry of the HCl and NH3 reaction. The balanced equation for the reaction is:
HCl + NH3 -> NH4Cl
From the equation, we can see that 1 mole of HCl reacts with 1 mole of NH3 to produce 1 mole of NH4Cl.
Since the number of moles of HCl in the final solution is 0.07, we'll need an equal number of moles of NH3 to neutralize it.
Therefore, the number of moles of NH3 needed = 0.07 moles.
(ii) To find the volume of a 2.0 mol dm^-3 NH3 solution needed to neutralize the final acid solution, we'll use the molarity and the number of moles.
Molarity of NH3 solution (M) = 2.0 mol dm^-3
Number of moles of NH3 needed = 0.07 moles
Volume (V) = Number of moles / Molarity
V = 0.07 / 2.0 = 0.035 dm^3
Since the volume is currently in dm^3, and 1 dm^3 = 1000 cm^3, we can convert the volume to cm^3:
V = 0.035 * 1000 = 35 cm^3
Therefore, 35 cm^3 of a 2.0 mol dm^-3 NH3 solution is needed to neutralize the final acid solution.