Calculate the pH of the resulting solution if 21.0 mL of 0.210 M HCl(aq) is added to
a) 26.0 mL of 0.210 M NaOH(aq)
b) 31.0 mL of 0.260 M NaOH(aq)
moles HCl = M x L = ??
moles NaOH = M x L = ??
Which is in excess.
delta moles = (H^+) or (OH^-) depending upon which in in excess, then
M = moles/total volume in L.
Finally, pH = =log(H^+) or pOH= -log(OH^-)
To calculate the pH of the resulting solution, we need to determine the moles of acid and base, and then use this information to calculate the concentration of the remaining species. Finally, we can use the concentration of hydroxide ions (OH-) to calculate the pOH, and then convert this value to pH.
a) Let's start with the reaction between HCl and NaOH:
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
First, let's calculate the moles of HCl and NaOH:
Moles of HCl = volume (L) × concentration (M)
= 0.021 L × 0.210 mol/L
= 0.00441 moles
Moles of NaOH = volume (L) × concentration (M)
= 0.026 L × 0.210 mol/L
= 0.00546 moles
Since HCl and NaOH react in a 1:1 ratio, the limiting reactant is HCl. Therefore, all of the HCl will react, leaving an excess of NaOH.
The moles of NaOH remaining = Moles of NaOH initially - Moles of HCl reacted
= 0.00546 moles - 0.00441 moles
= 0.00105 moles
Now, let's calculate the concentration of NaOH:
Concentration of NaOH = Moles of NaOH / volume (L)
= 0.00105 moles / 0.031 L
= 0.0339 M
Now, let's calculate the concentration of OH- ions:
Concentration of OH- = concentration of NaOH
= 0.0339 M
Next, we can calculate the pOH:
pOH = -log10(OH- concentration)
= -log10(0.0339)
≈ 1.47
Finally, we can calculate the pH:
pH = 14 - pOH
= 14 - 1.47
≈ 12.53
Therefore, the pH of the resulting solution when 21.0 mL of 0.210 M HCl(aq) is added to 26.0 mL of 0.210 M NaOH(aq) is approximately 12.53.
b) Following the same method as in part a, let's calculate the pH of the resulting solution when 21.0 mL of 0.210 M HCl(aq) is added to 31.0 mL of 0.260 M NaOH(aq):
Moles of HCl = 0.021 L × 0.210 mol/L = 0.00441 moles
Moles of NaOH = 0.031 L × 0.260 mol/L = 0.00806 moles
Moles of NaOH remaining = 0.00806 moles - 0.00441 moles = 0.00365 moles
Concentration of NaOH = 0.00365 moles / 0.031 L = 0.1177 M
Concentration of OH- = 0.1177 M
pOH = -log10(0.1177) ≈ 0.93
pH = 14 - pOH = 14 - 0.93 ≈ 13.07
Therefore, the pH of the resulting solution when 21.0 mL of 0.210 M HCl(aq) is added to 31.0 mL of 0.260 M NaOH(aq) is approximately 13.07.
To calculate the pH of a resulting solution after mixing acids and bases, we need to determine the concentration of the remaining acid or base and find the concentration of the resulting salt.
a) Let's start with the reaction between HCl and NaOH:
HCl(aq) + NaOH(aq) --> NaCl(aq) + H2O(l)
Given:
- Volume of HCl solution (VHCl) = 21.0 mL
- Concentration of HCl solution (CHCl) = 0.210 M
- Volume of NaOH solution (VNaOH) = 26.0 mL
- Concentration of NaOH solution (CNaOH) = 0.210 M
First, we need to determine how many moles of HCl and NaOH are present in their respective solutions.
Moles of HCl = CHCl * VHCl = 0.210 M * 0.021 L = 0.00441 mol
Moles of NaOH = CNaOH * VNaOH = 0.210 M * 0.026 L = 0.00546 mol
As per the balanced equation, the reaction between HCl and NaOH occurs in a 1:1 ratio. This means that all 0.00441 mol of HCl will react with an equal amount of NaOH.
Since NaOH is in excess, 0.00441 mol of NaOH will be completely consumed in the reaction, and the remaining amount is 0.00546 - 0.00441 = 0.00105 mol of NaOH.
Now, let's calculate the concentration of the remaining NaOH in the resulting solution:
Concentration of NaOH in the resulting solution = moles of NaOH / volume of resulting solution
Volume of resulting solution = VHCl + VNaOH = 0.021 L + 0.026 L = 0.047 L
Concentration of NaOH = 0.00105 mol / 0.047 L = 0.0223 M
To find the pH, we can use the equation for the hydroxide ion concentration (OH-):
pOH = -log10[OH-]
pOH = -log10[0.0223] = -(-1.65) = 1.65
Finally, we can find the pH using the equation:
pH = 14 - pOH
pH = 14 - 1.65 = 12.35
Therefore, the pH of the resulting solution is approximately 12.35.
b) The procedure for calculating the pH in part b) is the same, except we'll be using different volumes and concentrations:
Given:
- Volume of HCl solution (VHCl) = 21.0 mL
- Concentration of HCl solution (CHCl) = 0.210 M
- Volume of NaOH solution (VNaOH) = 31.0 mL
- Concentration of NaOH solution (CNaOH) = 0.260 M
Moles of HCl = CHCl * VHCl = 0.210 M * 0.021 L = 0.00441 mol
Moles of NaOH = CNaOH * VNaOH = 0.260 M * 0.031 L = 0.00806 mol
After the reaction, the remaining NaOH will be:
0.00806 - 0.00441 = 0.00365 mol
Volume of resulting solution = VHCl + VNaOH = 0.021 L + 0.031 L = 0.052 L
Concentration of NaOH in the resulting solution = 0.00365 mol / 0.052 L = 0.0702 M
pOH = -log10[0.0702] = -(-1.153) = 1.153
pH = 14 - pOH = 14 - 1.153 = 12.847
Therefore, the pH of the resulting solution is approximately 12.847.