Chemistry help please

Calculate the PH after the following solutions are mixed together:
a) 15cm3 of 0.1moldm-3 HCL and 10cm3 of 0.1moldm-3 NAOH. (The answer is 1.70 but i don't know how to work it out)

b) 10cm3 of 0.1moldm-3 HCL and 15cm3 of 0.1moldm-3 NAOH. (The answer is 12.30 but i don;t know how to work it out)

I find these questions difficlut because the Ka is not given. My teacher did not teach us how to work it out. I stumbled upon these same questions online, that is how i know the answer to them. HELP please! Thanks

Calculate the PH after the following solutions are mixed together:
a) 15cm3 of 0.1moldm-3 HCL and 10cm3 of 0.1moldm-3 NAOH. (The answer is 1.70 but i don't know how to work it out)

b) 10cm3 of 0.1moldm-3 HCL and 15cm3 of 0.1moldm-3 NAOH. (The answer is 12.30 but i don;t know how to work it out)

I find these questions difficlut because the Ka is not given. My teacher did not teach us how to work it out. I stumbled upon these same questions online, that is how i know the answer to them. HELP please! Thanks

NaOH + HCl > H20 + NaCl

So one mole of HCl neutralises one mole of NaOH.

a) 15/1000 x 0.1 = number on moles of HCl = 0.0015
10/1000 x 0.1 = number of moles of NaOH = 0.001
0.0015-0.001= 0.0005 = moles of HCl left over.
0.0005 x 1000/25 = Conc of HCl left over= 0.02
minus log based ten of 0.02 = 1.7

You're very welcome. Now do the second one yourself.

Also you only need Ka for partially ionised stuff. Strong acids are fully ionised.So the moles of acid (monoprotic) = moles of H+

These questions do not require ka values as the acids and bases are strong.

You must use the balanced equation to find out the excess reagent and calculate the pH based on the moles of acid or base remaining and the total volume...

Thanks you very much. You are a life saver.

Happy to help,PM me if you need get stuck.

NaOH + HCl > H20 + NaCl

So one mole of HCl neutralises one mole of NaOH.

a) 15/1000 x 0.1 = number on moles of HCl = 0.0015
10/1000 x 0.1 = number of moles of NaOH = 0.001
0.0015-0.001= 0.0005 = moles of HCl left over.
0.0005 x 1000/25 = Conc of HCl left over= 0.02
minus log based ten of 0.02 = 1.7

You're very welcome. Now do the second one yourself.

Also you only need Ka for partially ionised stuff. Strong acids are fully ionised.So the moles of acid (monoprotic) = moles of H+

can i get help in b part i am not being able to do please help anyone?

Check post 4, it explains clearly