What is the pH of the solution created by combining 11.20 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)?
NaOH + HCl ==> NaCl + HOH
strong base + strong acid.
Determine mols NaOH = M x L = ??
Determine mols HCl = M x L = ??
Determine which reactant is in excess, then calculate molarity NaOH or HCl and pH from that. Post your work if you get stuck.
To find the pH of the solution created by combining NaOH and HCl, we need to understand the concept of neutralization.
When a strong acid (HCl) reacts with a strong base (NaOH), they undergo a neutralization reaction, resulting in the formation of a salt (NaCl) and water (H2O).
First, we need to determine the amount of NaOH and HCl used in the solution. We are given that the volume of NaOH is 11.20 mL and its concentration is 0.10 M, and the volume of HCl is 8.00 mL and its concentration is also 0.10 M.
Next, we can calculate the moles of NaOH and HCl using the equation:
moles = volume (in liters) * concentration (in moles per liter)
For NaOH:
moles of NaOH = (11.20 mL / 1000 mL) * 0.10 M = 0.00112 moles
For HCl:
moles of HCl = (8.00 mL / 1000 mL) * 0.10 M = 0.00080 moles
Since NaOH and HCl react in a 1:1 mole ratio according to the balanced equation, we can see that HCl is the limiting reactant. This means that all of the HCl will react, and any excess NaOH will remain unreacted.
Now, we need to calculate the resulting concentration of NaCl, which is the only remaining ion affecting the pH. Since the moles of NaOH and HCl are equal, the moles of NaCl formed will also be equal to the moles of HCl used:
moles of NaCl = 0.00080 moles
To find the new solution's volume, we sum the initial volumes of NaOH and HCl:
total volume = 11.20 mL + 8.00 mL = 19.20 mL
Since the volume is given in mL, we need to convert it to liters:
total volume = 19.20 mL / 1000 mL = 0.0192 L
Finally, we can calculate the concentration of NaCl:
concentration of NaCl = moles of NaCl / total volume
concentration of NaCl = 0.00080 moles / 0.0192 L = 0.0417 M
The resulting solution's concentration of NaCl is 0.0417 M.
To calculate the pH of the solution, we need to understand that NaCl is a salt that dissociates into sodium ions (Na+) and chloride ions (Cl-), and these ions do not affect the pH of the solution since they are neither acidic nor basic. Therefore, the pH of the resulting solution will be determined by the presence of water.
The pH of pure water at room temperature is 7, which is considered neutral. Since the resulting solution contains only water and NaCl, which have no acidic or basic properties, the pH of the solution will also be 7.
Therefore, the pH of the solution created by combining 11.20 mL of 0.10 M NaOH(aq) with 8.00 mL of 0.10 M HCl(aq) is 7.