Strong acid reacting with a strong base and calculate the resulting pH.
Consider the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH):
HCl + NaOH → NaCl + H2O
This is a neutralization reaction, where the acid and base react to form salt and water. In this case, sodium chloride (NaCl) and water (H2O) are formed.
Since HCl is a strong acid and NaOH is a strong base, the reaction will go to completion, meaning all of the HCl and NaOH will react to form NaCl and H2O.
To calculate the resulting pH, we need to determine the concentration of H+ ions and OH- ions in the resulting solution. In a neutralization reaction like this, the moles of H+ ions will be equal to the moles of OH- ions, since the reaction goes to completion.
Let's say we start with 50 mL of 0.1 M HCl and 50 mL of 0.1 M NaOH. To determine the number of moles of each, we use the formula:
moles = concentration (M) x volume (L)
For HCl:
moles = 0.1 M x 0.05 L = 0.005 moles
For NaOH:
moles = 0.1 M x 0.05 L = 0.005 moles
Since the reaction goes to completion, all of the HCl and NaOH will react to form NaCl and H2O. Since the reaction produces 1 mole of water for every mole of HCl and NaOH, the number of moles of water produced will be equal to the number of moles of HCl and NaOH.
moles of water = moles of HCl + moles of NaOH = 0.005 moles + 0.005 moles = 0.01 moles
Therefore, the final volume of the solution will be 100 mL (50 mL HCl + 50 mL NaOH), and the concentration of H+ ions and OH- ions will be:
[H+] = [OH-] = moles of H+ ions/total volume
[H+] = [OH-] = 0.005 moles/0.1 L = 0.05 M
Now, to determine the pH, we use the formula:
pH = -log[H+]
pH = -log(0.05)
pH = 1.3
Therefore, the pH of the resulting solution is 1.3, which is highly acidic.
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