Question

calculate the PH of 1×10-9 molar solution of NaoH​

Answer 1

Answer:

Concentration of given solution is 10^-9 M

NaOH is a strong base. When dissolved in water, it quickly dissociates into Na+ and OH- ions.

So, concentration of [OH-] ions is 10^-9 M

Since the solution also contains water, concentration of [OH-] ions from water is 10^-7 M

Total concentration of [OH-] ions = (10^-7)+(10^-9) = 101*10^-9 M

pOH =-log[OH-]=-{ log (101*10^-9)} =-{log(101)+log(10^-9)} = -log(101)-log(10^-9) = 9-log(101) = 9–2.004 = 6.996

pH = 14-pOH = 14–6.996 = 7.004

Explanation:

$hope \: it \: helps \: uhh$