What is pH of an aqueous solution formed by adding 20 mL of 0.1 mol L-1 NaOH to 30 mL of 0.2 mol L-1 HCI?
Please show steps and maths. Is there a unit conversion if so how is this done.
Thank you
HCl + NaOH ==> NaCl + H2O
moles HCl initially = M x L = ??
moles NaOH initially = M x L = ??
Since these react, one of them will be used completely and the other will have some remaining. What you do is subtract the smaller moles of reagent from the larger moles of reagent, then use the excess of the remaining solution (still in moles) to convert to M (M = moles/L), then to pH.
To find the pH of the aqueous solution formed by mixing NaOH and HCl, we need to determine the concentration of the resulting solution. This can be done using the concept of stoichiometry.
Step 1: Convert the given volumes to liters:
20 mL = 20/1000 = 0.02 L (for NaOH)
30 mL = 30/1000 = 0.03 L (for HCl)
Step 2: Calculate the moles of NaOH and HCl:
Moles of NaOH = Volume (in L) x Concentration (mol L^-1)
= 0.02 L x 0.1 mol L^-1
= 0.002 mol
Moles of HCl = Volume (in L) x Concentration (mol L^-1)
= 0.03 L x 0.2 mol L^-1
= 0.006 mol
Step 3: Determine the limiting reactant:
Since NaOH and HCl react in a 1:1 ratio based on their balanced equation, the limiting reactant is the one with fewer moles. In this case, it is NaOH, with 0.002 mol.
Step 4: Determine the excess reactant:
Since NaOH is the limiting reactant, all of it reacts completely with HCl. Thus, there is no excess of HCl remaining.
Step 5: Calculate the final volume of the solution:
The total volume of the solution is the sum of the initial volumes of NaOH and HCl:
Total volume = 20 mL + 30 mL = 50 mL = 50/1000 = 0.05 L
Step 6: Calculate the concentration of the resulting solution:
The moles of NaOH react completely with an equal number of moles of HCl in the final solution, i.e., 0.002 mol.
Therefore, the concentration of the resulting solution = moles/volume = 0.002 mol/0.05 L = 0.04 mol L^-1.
Step 7: Calculate the pOH of the resulting solution:
pOH = -log10 [OH-]
Since NaOH is a strong base, it dissociates completely in water, yielding 1 mole of OH- ions for every mole of NaOH. Therefore, [OH-] = 0.002 mol L^-1.
pOH = -log10 (0.002)
= -log10 (2 x 10^-3)
= -log10 (2) - (-log10 (10^3))
Using the values from the logarithmic table, -log10 (2) is approximately 0.30103 and log10 (10^3) is 3.
Therefore, pOH ≈ 0.30103 + 3
= 3.30103
Step 8: Calculate the pH of the resulting solution:
pH + pOH = 14 (at 25°C)
pH = 14 - pOH
= 14 - 3.30103
= 10.69897
Therefore, the pH of the aqueous solution formed by mixing 20 mL of 0.1 mol L^-1 NaOH and 30 mL of 0.2 mol L^-1 HCl is approximately 10.69897.
Unit conversion:
In this problem, we converted the given volumes of NaOH and HCl from milliliters (mL) to liters (L). This conversion was done by dividing the given volume by 1000, as there are 1000 milliliters in a liter.