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
2 answers
what are you solving for? pH or H+
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.
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.
1 answer