Question

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

Answer 1

what are you solving for? pH or H+

Answer 2

<b>Thanks for showing your work. It makes it easy to find the error. See below</b>

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)
<b>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.</b>
[H+] = 1.309 × 10^-4 M or 1.3 × 10^-4 M
<b>My answer is something like 0.0108 for pH = 1.965 which I would round to 1.96.</b>