what is the pH of the solution formed when 25 mL of 0.173 M NaOH is added to 35 mL of 0.342 M HCl?
millimols HCl = 35 mL x 0.342M = ?
mmols NaOH = 5 x 0.173 = ?
Calculate each, determine the one in excess, calculate the amount in excess, then if acid pH = -log(H^+) or if basic then pOH = -log(OH^-) and convert to pH by pH + pOH = pKw = 14.
To find the pH of the solution formed when NaOH and HCl react, we need to determine the amount of excess HCl or NaOH remaining after the reaction has occurred. With this information, we can compute the concentration of the remaining acid or base and use it to calculate the pH.
First, let's determine the number of moles of NaOH and HCl by using the equation:
moles = concentration × volume
Moles of NaOH = 0.173 M × 0.025 L = 0.004325 mol
Moles of HCl = 0.342 M × 0.035 L = 0.01197 mol
Since HCl and NaOH react in a 1:1 ratio according to the balanced chemical equation (HCl + NaOH -> NaCl + H2O), we can see that HCl is in excess, leaving NaOH as the limiting reagent.
Next, we need to determine the moles of HCl that reacted with NaOH. Since the reaction is 1:1, the moles of HCl that reacted with NaOH are also 0.004325 mol.
Now, let's calculate the moles of HCl remaining after the reaction:
Moles of HCl remaining = initial moles of HCl - moles of HCl reacting with NaOH
= 0.01197 mol - 0.004325 mol
= 0.007645 mol
To find the concentration of HCl remaining, we divide the moles of HCl remaining by the total volume of the solution:
Concentration of HCl remaining = moles of HCl remaining / total volume of solution
= 0.007645 mol / (0.025 L + 0.035 L)
≈ 0.0987 M
Now that we have the concentration of HCl remaining, we can find the pH of the solution using the formula:
pH = -log[H+]
[H+] represents the concentration of H+ ions in the solution. Since HCl is a strong acid, it fully dissociates in water, meaning the concentration of H+ ions is equal to the concentration of HCl.
Therefore, the pH can be calculated as:
pH = -log(0.0987)
≈ 1.01
Hence, the pH of the solution formed when 25 mL of 0.173 M NaOH is added to 35 mL of 0.342 M HCl is approximately 1.01.