Five solutions A, B, C, D and E when tested with universal indicator showed pH as 4, 1, 11, 7 and 9 respectively, which solution is:
(a) neutral (b) strongly alkaline (c) strongly acidic (d) weakly acidic (e) weakly alkaline Arrange the pH in increasing order of hydrogen ion concentration.

Five solutions A, B, C, D, E, F as 5, 2, 1, 3, 7 and 9 respectively which solution is (a) Neutral (b) Strongly alkaline (c ) Strongly acid (d) Weakly acidic Arrange the pH in increasing order of Hydrogen ion concentration.

You have two solutions, A and B. The pH of solution A is 6 and pH of solution B is 8. Which solution has more hydrogen ion concentration? Which of this is acidic and which one is basic?

The results of reactions of metals A, B, C, D, E with different solutions are given in the table below. Observe the table and write answers. Arrange the metals A, B, C, D, E in the order of their reactivity ?

When a salt reacts with water to form acidic (or ) basic solution the process is called salt hydrolysis. The P^(H) of salt solution can be calculated using the following relation P^(H) =(1)/(2) [P^(k_w) +P^(Ka) + log C ] for salt of weak acid and strong base. P^(H) =(1)/(2) [P^(K_w) -P^(K_a) -log C ] for salt of weak base and strong acid , P^(H) =(1)/(2) [P^(K_w) +P^(K_a) - P^(K_b) ] For a salt of weak acid and weak base where .c. represents the concentration of salt . When a weak acid (or) a weak base is not completly neatralised by strong base (or ) strong acid respectively, then formation of buffer takes places. The P^(H) of buffer solution can be calculated using the following relation P^(H) =P^(Ka) +log "" ((["salt")])/(["Acid "]) , P^(OH) =P^(Kb) +log ""(["salt"])/(["base"]) Answer the following questions using the following data pK_a (CH_3COOH) =4.7447 , pK_b (NH_4OH) =4.7447 , P^(K_W) =14. One mole CH_3 COOH and one mole CH_3 COONa are dissolved in water one litre aqueous solution The P^(H) of the resulting solution will be

When a salt reacts with water to form acidic (or ) basic solution the process is called salt hydrolysis. The P^(H) of salt solution can be calculated using the following relation P^(H) =(1)/(2) [P^(k_w) +P^(Ka) + log C ] for salt of weak acid and strong base. P^(H) =(1)/(2) [P^(K_w) -P^(K_a) -log C ] for salt of weak base and strong acid , P^(H) =(1)/(2) [P^(K_w) +P^(K_a) - P^(K_b) ] For a salt of weak acid and weak base where .c. represents the concentration of salt . When a weak acid (or) a weak base is not completly neatralised by strong base (or ) strong acid respectively, then formation of buffer takes places. The P^(H) of buffer solution can be calculated using the following relation P^(H) =P^(Ka) +log "" ((["salt")])/(["Acid "]) , P^(OH) =P^(Kb) +log ""(["salt"])/(["base"]) Answer the following questions using the following data pK_a (CH_3COOH) =4.7447 , pK_b (NH_4OH) =4.7447 , P^(K_W) =14. 0.001 M NH_4 Cl aqueous solution has P^(H)

An organic compound A(C_4 H_7 C_(13)) yields (B) when treated with aq. KOH. (B) upon treatment with C_2 H_5 OH in presence of acid gave (C) which upon reducing with LiAIH_4 gave (D) and (E). (B) upon treatment with NH_3 followed by heating with P_4 O_(10) and subsequent hydrolysis gives back (B). Sodium salt of (B) on Kolbe's electrolysis gave 2, 3-dimethylbutane at anode. The dehydration of compound (D) gives