how can i calculate the pH of the solution obtained when 50.00mL of 0.150mol/L hydrochloric acid is added to 75mL of a 0.111 mol/L aqueous sodium hydroxide solution.
need help i am stuck with that question thank you :)
HCl + NaOH ==> NaCl + H2O
moles HCl = M x L = ??
moles NaOH = M x L = ??
determine which is in excess and calculate H^+ (if HCl is in excess) or OH^- (if NaOH is in excess), and pH from there.
To calculate the pH of the solution, we need to determine the concentration of hydroxide ions (OH-) and hydrogen ions (H+). Since hydrochloric acid is a strong acid and sodium hydroxide is a strong base, they dissociate completely in water.
First, let's determine the concentration of hydroxide ions (OH-) in the final solution using the equation:
(M1)(V1) = (M2)(V2)
Where:
M1 = initial concentration of sodium hydroxide
V1 = initial volume of sodium hydroxide
M2 = final concentration of hydroxide ions
V2 = final volume of the solution
Given:
M1 = 0.111 mol/L
V1 = 75 mL
M2 = ?
V2 = (50 mL + 75 mL) = 125 mL
Now, let's plug in the values and solve for M2:
(0.111 mol/L)(75 mL) = (M2)(125 mL)
0.008325 mol = (M2)(125 mL)
M2 = 0.008325 mol / 125 mL
M2 = 0.0666 mol/L
Therefore, the concentration of hydroxide ions (OH-) in the final solution is 0.0666 mol/L.
Next, since hydrochloric acid is a strong acid, it dissociates fully into hydrogen ions (H+). Therefore, the concentration of hydrogen ions (H+) is equal to the concentration of hydrochloric acid, which is 0.150 mol/L.
Now, we can calculate the pH using the equation:
pH = -log[H+]
Where [H+] is the concentration of hydrogen ions. Plugging in the value, we get:
pH = -log(0.150)
pH = -(-0.8239)
pH = 0.8239
Therefore, the pH of the solution obtained when 50.00 mL of 0.150 mol/L hydrochloric acid is added to 75 mL of a 0.111 mol/L aqueous sodium hydroxide solution is approximately 0.8239.