Find the pH of a solution obtained by mixing 100cm3 of 01MHCl solution with 100cm3 of 0.2M NaOH
To find the pH of the solution, we need to determine the concentration of H+ ions in the final solution. First, let's calculate the moles of H+ in both the HCl and NaOH solutions.
Moles of H+ in HCl solution:
Molarity (M) = moles/volume (L)
0.1 M = moles/0.1 L
moles of H+ in HCl solution = 0.1 M * 0.1 L = 0.01 moles
Moles of OH- in NaOH solution:
Molarity (M) = moles/volume (L)
0.2 M = moles/0.1 L
moles of OH- in NaOH solution = 0.2 M * 0.1 L = 0.02 moles
Since HCl is a strong acid and NaOH is a strong base, they will react in a 1:1 ratio to form water (H2O) and a neutral salt, NaCl.
HCl + NaOH → H2O + NaCl
Since the moles of H+ and OH- are equal, they will neutralize each other completely, leaving no excess H+ or OH- ions in the solution.
Therefore, the final solution will contain 0.01 moles of NaCl and no excess H+ or OH- ions.
To calculate the pH, we'll convert the concentration of H+ ions into pH using the formula:
pH = -log[H+]
Since the concentration of H+ ions is equal to the concentration of OH- ions, we can calculate the concentration of H+ ions using the moles of HCl:
Concentration of H+ ions = moles/volume (L)
Concentration of H+ ions = 0.01 moles/0.2 L = 0.05 M
Now, we can find the pH:
pH = -log[H+]
pH = -log(0.05)
pH ≈ -(-1.301)
pH ≈ 1.301
Therefore, the pH of the solution obtained by mixing 100 cm3 of 0.1 M HCl solution with 100 cm3 of 0.2 M NaOH is approximately 1.301.
To find the pH of the solution obtained by mixing 100 cm3 of 0.1 M HCl solution with 100 cm3 of 0.2 M NaOH, we need to determine the concentration of the resulting solution.
Step 1: Determine the amount of moles of HCl and NaOH used.
Since concentration (C) is equal to the number of moles (n) divided by the volume (V), we can use the formula:
n = C * V
For 100 cm3 of 0.1 M HCl, the number of moles is:
n(HCl) = 0.1 mol/L * 0.1 L = 0.01 mol
For 100 cm3 of 0.2 M NaOH, the number of moles is:
n(NaOH) = 0.2 mol/L * 0.1 L = 0.02 mol
Step 2: Determine the limiting reactant.
To find the pH, we need to determine the excess reactant. The reactant that is not fully consumed is referred to as the limiting reactant.
The balanced chemical equation for the reaction between HCl and NaOH is: HCl + NaOH -> NaCl + H2O
From the balanced equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH. Therefore, the limiting reactant will be the one for which there is an insufficient amount to react with the other completely.
In this case, since we have equal moles of HCl and NaOH (0.01 mol and 0.02 mol, respectively), HCl is the limiting reactant because it will be completely consumed while NaOH will have an excess.
Step 3: Determine the moles of the excess reactant remaining.
Since HCl is the limiting reactant, all of it will react with NaOH. Therefore, the concentration of HCl in the final solution will become zero. The excess reactant, NaOH, will remain in the solution.
The moles of remaining NaOH can be calculated by subtracting the moles of HCl consumed from the initial moles of NaOH:
moles of remaining NaOH = initial moles of NaOH - moles of HCl consumed
= 0.02 mol - 0.01 mol
= 0.01 mol
Step 4: Determine the concentration of the final solution.
The total volume of the resulting solution is the sum of the volumes of HCl and NaOH used, which is 100 cm3 + 100 cm3 = 200 cm3.
To convert the volume to liters, we divide it by 1000:
V = 200 cm3 / 1000 cm3/L = 0.2 L
The concentration of the final solution can be determined using the equation:
C_final = moles of excess reactant / volume of final solution
Therefore, the concentration of the final solution is:
C_final = 0.01 mol / 0.2 L
C_final = 0.05 M
Step 5: Calculate the pH of the final solution.
To find the pH of the solution, we need to use the concentration of hydroxide ions (OH-) present in the solution. Since we have NaOH in excess, the concentration of hydroxide ions can be calculated using the concentration of the final solution and the volume of NaOH used (100 cm3 or 0.1 L):
OH- concentration = NaOH concentration = 0.2 M
To calculate the pOH, we can use the formula:
pOH = -log10 [OH- concentration]
pOH = -log10 (0.2)
pOH ≈ 0.7
Since pH + pOH = 14, we can determine the pH of the solution by subtracting the pOH value from 14:
pH = 14 - pOH
pH = 14 - 0.7
pH ≈ 13.3
Therefore, the pH of the solution obtained by mixing 100 cm3 of 0.1 M HCl with 100 cm3 of 0.2 M NaOH is approximately 13.3.
Convert volume to L
100cm^3=100mL=0.100L
Multiply your volume by the molarity given for each one and subtract moles of acid from moles of base.
looking at the numbers, I see that you are going to get more moles of base than acid, (i.e., 0.01 moles of NaOH. But double check.
molarity of unreacted NaOH=(0.01 moles of NaOH/total volume of acid and base used (0.2L)
plug in the value obtain from above in the below equation and solve for pH.
pOH=-log[OH-]
pH=14-pOH