What is the pH of the solution obtained by mixing 30.00mL OF 0.250M HCL AND 30.00mL OF 0.125M NaOH?
To find the pH of the solution obtained by mixing 30.00 mL of 0.250 M HCl and 30.00 mL of 0.125 M NaOH, you need to calculate the concentrations of the resulting solution's hydronium ions (H3O+) or hydroxide ions (OH-).
Step 1: Determine the moles of HCl and NaOH used.
Moles (HCl) = concentration (HCl) x volume (HCl)
Moles (HCl) = 0.250 M x 0.0300 L = 0.0075 moles
Moles (NaOH) = concentration (NaOH) x volume (NaOH)
Moles (NaOH) = 0.125 M x 0.0300 L = 0.00375 moles
Step 2: Determine the limiting reagent.
Since NaOH and HCl react in a 1:1 ratio, the limiting reagent is the one with fewer moles. In this case, NaOH is the limiting reagent because it has fewer moles.
Step 3: Determine the excess moles of the excess reagent.
Excess moles (HCl) = moles (HCl) - moles (NaOH)
Excess moles (HCl) = 0.0075 moles - 0.00375 moles = 0.00375 moles
Step 4: Determine the moles of water formed in the reaction.
Moles of water formed = moles of limiting reagent = moles (NaOH) = 0.00375 moles
Step 5: Determine the concentration of hydronium ions (H3O+) in the solution.
The number of moles of hydronium ions formed is equal to the number of moles of water formed.
Concentration (H3O+) = moles (H3O+) / volume of solution
Volume of solution = volume (HCl) + volume (NaOH) = 0.0300 L + 0.0300 L = 0.0600 L
Concentration (H3O+) = 0.00375 moles / 0.0600 L = 0.0625 M
Step 6: Calculate the pH of the solution using the formula:
pH = -log10(concentration of H3O+)
pH = -log10(0.0625)
pH ≈ 1.20
Therefore, the pH of the solution obtained by mixing 30.00 mL of 0.250 M HCl and 30.00 mL of 0.125 M NaOH is approximately 1.20.
To find the pH of the solution obtained by mixing 30.00 mL of 0.250 M HCl and 30.00 mL of 0.125 M NaOH, we can use the concept of acid-base neutralization.
Step 1: Determine the moles of HCl and NaOH:
Moles of HCl = (Volume of HCl in liters) x (Molarity of HCl)
= (0.03000 L) x (0.250 mol/L)
= 0.00750 mol
Moles of NaOH = (Volume of NaOH in liters) x (Molarity of NaOH)
= (0.03000 L) x (0.125 mol/L)
= 0.00375 mol
Step 2: Determine the limiting reactant:
Since HCl and NaOH react in a 1:1 ratio (1 mol HCl reacts with 1 mol NaOH), the limiting reactant will be whichever is present in lower moles. In this case, NaOH is the limiting reactant because there are fewer moles of NaOH (0.00375 mol) compared to HCl (0.00750 mol).
Step 3: Determine the excess reactant:
Since HCl is in excess, whatever amount of HCl is not used in the neutralization will determine the composition of the solution. We will need to calculate the moles of HCl that are in excess.
Moles of HCl in excess = Moles of HCl - Moles of NaOH
= 0.00750 mol - 0.00375 mol
= 0.00375 mol
Step 4: Determine the concentration of the excess HCl in the final solution:
To determine this, we need to calculate the total volume of the solution after the 30.00 mL HCl and 30.00 mL NaOH are mixed.
Total volume of the solution = Volume of HCl + Volume of NaOH
= 0.03000 L + 0.03000 L
= 0.06000 L
Concentration of the excess HCl in the final solution = (Moles of HCl in excess) / (Total volume of the solution in liters)
= 0.00375 mol / 0.06000 L
≈ 0.0625 M
Step 5: Calculate the pH of the solution:
The pH can be determined using the equation: pH = -log[H+].
[H+] can be determined from the concentration of H+ ions in the excess HCl, which is equal to the concentration of the excess HCl.
pH = -log(0.0625)
≈ 1.20
Therefore, the pH of the solution obtained by mixing 30.00 mL of 0.250 M HCl and 30.00 mL of 0.125 M NaOH is approximately 1.20.
Write the equation and balance it.
NaOH + HCl ==> NaCl + H2O
Calculate moles NaOH.
Calculate moles HCl.
Set up an ICE chart to see what you finish with.
If the moles base and moles acid are the same, they exactly neutralize each other and the pH will be exactly neutral for you will have table salt in water. If the base is in excess, calculate OH from that. If the acid is in excess, calculate H^+ from that.