Calculate the pH of a mixture of 20.0 mL of 0.150 M HCl and 10.0 mL of 0.300 M NaOH.​[7.00]
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20ml(0.150M HCl) + 10ml(0.0300M NaOH
=> 0.003mole HCl + 0.003mole NaOH
=> (0.003mol/0.03L)HCl + (0.003mol/0.01L)NaOH
=> 0.30M HCl + 0.30M NaOH
=> 0.30M NaCl + 0.30M HOH
Since neither Na^+ or Cl^- will undergo hydrolysis (rxn with water), the outcome of the reaction is dependent upon the autoionization of water only. That is,
HOH <=> H^+ + OH^-
=> [H^+] = [OH^-] = 1 x 10^-7
=> pH = -log[H^+] = -log(1 x 10^-7) = 7.00
To calculate the pH of the mixture, we will need to consider the reaction that occurs between HCl (a strong acid) and NaOH (a strong base). The balanced chemical equation for this reaction is:
HCl + NaOH -> NaCl + H2O
From this equation, we can see that HCl reacts with NaOH to form water and sodium chloride (NaCl).
To calculate the pH of the mixture, we need to determine the amount of excess HCl or NaOH that remains after the reaction. In this case, we have 20.0 mL of 0.150 M HCl and 10.0 mL of 0.300 M NaOH.
First, let's determine the number of moles of HCl and NaOH:
moles of HCl = volume of HCl (in L) x concentration of HCl (in mol/L)
= 0.020 L x 0.150 mol/L
= 0.003 mol
moles of NaOH = volume of NaOH (in L) x concentration of NaOH (in mol/L)
= 0.010 L x 0.300 mol/L
= 0.003 mol
Since the stoichiometry of the reaction is 1:1 between HCl and NaOH, both compounds completely react. Therefore, there is no excess HCl or NaOH left after the reaction.
Now, let's calculate the amount of water formed:
moles of water = minimum(moles of HCl, moles of NaOH)
= minimum(0.003 mol, 0.003 mol)
= 0.003 mol
Since the solution is now only water and sodium chloride, we can calculate the concentration of OH- ions in the solution using the moles of water formed:
OH- concentration = moles of water / total volume of solution (in L)
= 0.003 mol / (20.0 mL + 10.0 mL)
= 0.003 mol / 0.030 L
= 0.100 M
Now, let's calculate the pOH of the solution:
pOH = -log(OH- concentration)
= -log(0.100)
= 1.00
Finally, we can calculate the pH of the solution using the equation:
pH = 14.00 - pOH
= 14.00 - 1.00
= 13.00
Therefore, the pH of the mixture is 13.00.