Question

What is the pH when 30 ml of 1M hydrochloric acid is added to 200 mL of distilled water?

please writing hot to solve it .

Answer 1

Molarity of HCl = 1 M
Volume of HCl = 30 mL = 30/1000 L = 0.03 L
Number of moles of HCl = Molarity * Volume = 1M * 0.03L = 0.03 mol

Now after addinng 200 mL of water, Volume becomes = 30mL+ 200 mL
                                                    = 230 mL = 230/1000 L = 0.23 L

So,
New Molarity of HCl = number of moles / new Volume
                          = 0.03 / 0.23
                          =0.13 M

[H+] = 0.13 M
pH = -log [H+]
   = - log (0.13)
   = 0.88

Answer: 0.88