Sandu

1.025g of an impure sample of a weak acid HX(mol mass = 82) is dissolved in 60ml of water and titrated with 0.25M NaOH. When half of the acid was neutralized the pH was found to be 5.0 and at the equivalence point the pH is 9.0. Find the % purity of this...

How many grams of solid KOH must be added to 100ml of a buffer solution which is 0.1M each with respect to the weak acid HA and its potassium salt KA in order to make the pH of the solution 6.0?

In the titration of a solution of a weak acid HA with NaOH the pH is 5.0 after 10ml of NaOH solution has been added and 5.6 after 20ml of the NaOH solution has been added. Find the PKa of HA? Pls give some hints to solve this question...

To what volume should 10L of 0.5M CH3COOH (Ka = 1.8×10–5) be diluted in order to double the OH– concentration?

Find the ratio of the pH of solution (I) containing 1 mol of CH3COONa and 1 mol of HCl in 1L and solution (II) containing 1 mol of CH3COONa and 1 mol of CH3COOH in 1L

Thank you DrBob222... Got the answer as 80% purity.

Yes... Got the answer 0.458 g.. Thank you...

Normality = No. of gram equivalents of H2SO4/ volume of solution (litres) = (27/49) ------- 100×10^-3 = 5 Normal

Thank you DrBob222 for the correction..

Can we use Henderson -Hasselbalch equatiin for acid buffer to solve this probkem...??

Yes DrBob222... Thanks for your response... I tried previous problem as given below.. Don't know whether correct... Let 'a' be initial milli moles of HA and molarity of NaOH be 'x' In both case solution is acidic.. ie acidic buffer HA + NaOH --> NaA +H2O...

5 = Pka+ log 10x/a-10x -----(1) 5.6 = Pka+ log 20x/a-20x -----(2) (2) - (1) --> 0.6= log(20x/a-20x)- log(10x/a-10x) Since Pka is cancelled during subtraction. Solving the above eqn taking RHS in the form log a - log b = log a/b We get a = 30x Substituting...

How the concentration of HAc in first solution became 0.5M Can you explain pls