What is the pH when 1mL of 0.1N HCl is added to neutral solution?

Answer 1

No. of moles of HCl=molarity x volume=250x30=1500 milli moles

No. of moles of NaOH=molarity x volume=125x30=750 milli moles

firstly a neutralisation reaction takes place between HCl and NaOh

HCl(750millimoles)+NaOH(750millimoles)-------NaCl(750millimoles)+H2O

as the no. of Millimoles of base is less than that of the acid so the base is not completely neutralised and in the solution we have NaCl(750millimoles) and HCl(750millimoles)left.

as HCl Is a strong acid and NaCl is neutral hence the overall solution is Strongly basic.

concentraation of H+ is 60 ml

molarity=750/60=125M

pH= -log [H+]

pH=0,highly acidic solution

Answer 2

Since pH is a measurement value of concentration of the strong proton donor (HCl, acid), it depends on to how many liters neutral solute it is added to:

The so-called dilution factor determines the concentration of the protons and thus the pH value.

• pH of 1 ml 0.1 M HCl = 1.0
• This added to 9 ml water (10-fold dilution) will raise pH 1.0 unit (remember the logarithmic scale of pH!) so pH=2.0
• If added to 99 ml >> 100x dilution >> pH=3.0
• .... to 999 ml >> 1000x >> pH=4.0
• This goes on till the concentrations of protons reach a value of 10-6 (pH=6.0)

  At lower concentrations of acid the 'internal' proton concentration of the added water itself (= 1.0*10-7) becomes more and more significant in the resulting pH value. Thus it will never become greater than pH=7 by simply diluting it with water.

Answer 3

To calculate the pH of the solution, you need to know the [H3O+] because the pH is the -log [H3O+]. This can be determined from the [OH-] because [H3O+] x [OH-] = 1x10^-14.

Also, pH + pOH = 14

2.00 g NaOH x 1 mol NaOH/40 g = 0.05 moles NaOH

0.05 moles NaOH/0.075 L = 0.667 mole/L = [OH-]

pOH = -log [OH-] = -log 0.667 = 0.176

pH = 14 - 0.176 = 13.8

Using [H3O+][OH-] = 1x10^-14 and solving for [H3O+].....

1x10^-14/0.667 = [H3O+]

[H3O+] = 1.5 x10^-14 M

pH = -log 1.5x10^-14 = 13.8

Answer 4

11.80