Advertisements
Advertisements
प्रश्न
Calculate the pH of 1.5 × 10−3 M solution of Ba(OH)2.
Advertisements
उत्तर
\[\ce{\underset{1.5×10^{-3}M}{Ba(OH)2} -> Ba^{2+} + \underset{2×1.5×10^{-3}M}{2OH^-}}\]
[OH–] = 3 × 10−3 M
[∵ pH + pOH = 14]
pH = 14 – pOH
pH = 14 – (– log [OH–])
= 14 + log [OH–]
= 14 + log (3 × 10−3)
= 14 + log 3 + log 10−3
= 11 + 0.4771
pH = 11.48
APPEARS IN
संबंधित प्रश्न
The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization constant and pKb.
Calculate the pH of the resultant mixtures: 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl.
Define pH.
Calculate the pH and pOH of 0.0001 M HCl solution.
The least basic hydroxide from the following is:
The aqueous solution of sugar does not conduct electricity. However, when sodium chloride is added to water, it conducts electricity. How will you explain this statement on the basis of ionisation and how is it affected by concentration of sodium chloride?
Assertion (A): Increasing order of acidity of hydrogen halides is \[\ce{HF < HCl < HBr < HI}\]
Reason (R): While comparing acids formed by the elements belonging to the same group of periodic table, \[\ce{H - A}\] bond strength is a more important factor in determining acidity of an acid than the polar nature of the bond.
Derive the relationship between pH and pOH.
Define pH.
Derive the relation pH + pOH = 14.
