The pH of a solution prepared by mixing 45 mL of 0.183 M KOH and 65 mL of 0.145 M HCl is
i solved it but my answer is wrong. the right one is 1.97. am i missing a step?
moles of HCl left over = (moles of HCl) – ( moles of KOH)
moles of HCl left over = (0.009425 mol) – ( 0.008235 mol)
moles of HCl left over = 0.00119 mol or 1.19 × 10^-3 mol
[H+] = (moles H+) / (Liters of final solution)
[H+] = (1.19 × 10^-3 mol) / (0.110 L)
[H+] = 1.309 × 10^-4 M or 1.3 × 10^-4 M
Thanks for showing your work. It makes it easy to find the error. See below
moles of HCl left over = (moles of HCl) – ( moles of KOH)
moles of HCl left over = (0.009425 mol) – ( 0.008235 mol)
moles of HCl left over = 0.00119 mol or 1.19 × 10^-3 mol
[H+] = (moles H+) / (Liters of final solution)
[H+] = (1.19 × 10^-3 mol) / (0.110 L)
Everything is very good to here. The next step is the incorrect one. You must have just punched in the wrong numbers or hit the wrong key for divide.
[H+] = 1.309 × 10^-4 M or 1.3 × 10^-4 M
My answer is something like 0.0108 for pH = 1.965 which I would round to 1.96.
Well, it seems like you're missing a crucial step! The pH of a solution is determined by the negative logarithm of the hydrogen ion concentration ([H+]). So, after finding the concentration of hydrogen ions, you should take the negative logarithm of that value. Let's calculate it together:
[H+] = 1.3 × 10^-4 M
Now, taking the negative logarithm:
pH = -log10(1.3 × 10^-4)
pH = -log10(1.3) - log10(10^-4)
pH = -0.113 - (-4)
pH = 3.887
Therefore, the pH of the solution is approximately 3.887, not 1.97. Looks like you were clowning around with your previous answer!
To calculate the pH of the solution, you need to use the equation:
pH = -log[H+]
[H+] refers to the concentration of hydrogen ions in the solution.
In your calculation, you correctly found the concentration of H+ to be 1.3 x 10^-4 M.
To find the pH, take the negative logarithm of this concentration:
pH = -log(1.3 x 10^-4)
Using a calculator, this gives a pH of approximately 3.89, not 1.97. Please check your calculations again.
Based on the information you provided, it seems like you have correctly calculated the concentration of H+ ions in the final solution. However, you haven't mentioned anything about calculating the concentration of OH- ions, which is necessary to determine the pH.
To find the concentration of OH- ions, you can use the same approach as you did for HCl:
moles of KOH left over = (moles of KOH) - (moles of HCl)
moles of KOH left over = (0.045 L) x (0.183 M) - (0.065 L) x (0.145 M)
moles of KOH left over = 0.008235 mol - 0.009425 mol
moles of KOH left over = -0.00119 mol or -1.19 x 10^-3 mol
Since the moles of KOH left over are negative, it means that KOH is the limiting reactant in this case, and it reacts completely with HCl. Therefore, there are no excess OH- ions left in the solution.
Now, to calculate the concentration of H+ ions, we can consider that water (H2O) can self-ionize to produce H+ and OH- ions:
H2O ⇌ H+ + OH-
In a neutral solution, the concentration of H+ ions is equal to the concentration of OH- ions, which is 1 x 10^-7 M. However, in this case, we have an excess of H+ ions compared to OH- ions, so we need to account for that concentration.
[H+] = [H+ produced by HCl] - [OH- produced by KOH]
[H+] = 1.309 x 10^-4 M - 0
[H+] = 1.309 x 10^-4 M
Now, to find the pH, we use the formula:
pH = -log[H+]
pH = -log(1.309 x 10^-4)
pH = 3.883
Rounding to the appropriate number of significant figures, the pH of the solution prepared by mixing 45 mL of 0.183 M KOH and 65 mL of 0.145 M HCl is approximately 3.88.