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The dissociation constant of a substituted benzoic acid at $ 25 ^\circ C is 1.0 \times 10 ^{-4}. The P^H $ of 0.01 M solution of its sodium salt is
$ P ^ H = + { 1 \over 2 } ( P ^K W + P ^K a + log C ) $ $ = 1/2 \times 14 + 1/2 \times 4 + 1 /2 log 10 ^ {-2} $ = 7+2 -1 = 8
Number of $ H ^+ ions present in 500 ml of lemon juice of P ^ H = 3 $ is
$ P^H = 3 means [H^+] = 10 ^{-3} M $ 1000 ml juice contains $ 10 ^ {-3 }mole H ^ + $ ions $ \therefore no.of H ^ + ions = 10 ^{-3} \times 6.022 \times 10^{23} $ in 1000 ml $ 500 ml juice contains H^+ ions = { 10 ^{-3 } \times 6.022 \times 10^{23} \times 500 \over 1000} $ $ = 3.011 \times 10 ^ {20} $
Equimolar solution of the following were prepared in water separately. Which one of the solutions will record the highest $ P^H $ (?)
All alkaline earthmetalchlorides $ MCl_2$ on hydrolysis will produce acidic solution $ MCl_2 + H_2O \rightleftharpoons M(OH)_2 + 2HCl $ because $ M(OH)_2$ is a weak base and HCl is a strong acid. but as we go down the group, basic character of hydroxides increses. Hence acidic character decreses. So $ BaCl_2 will have the highest P^H$ .
Solubility products constants (KSP) of the salt types $ MX, MX_2 and M_3X at temp T. are 4 \times 10 ^{-8} , 3.2 \times 10 ^{-14 } and 2.7 \times 10 ^{-15} $ respectively. Solubility of the salts at temp. T are in the order,
$ MX_{(S)} \rightleftharpoons M ^ + + X ^ - $ $ K_{SP} = S^2 \therefore S = (K_{SP}) ^ {1/2} = (4 \times 10^ {-8} ) ^ {1/2} = 2 \times 10 ^ {-4} M $ $ MX_{2(s)} \rightleftharpoons M^{2+} + 2 X ^ - K_{SP} = 4 S ^3 \therefore S = ( { K_{SP} \over 4 } ) ^ {1/3} = 2 \times 10 ^ {-5} M $ $ M_3 X_{(s)} \rightleftharpoons 3M^+ + X^ {-3} $ $ K_{SP} = 27 S^4 \therefore S = ( { K_{SP} \over 27 } ) ^ {1/4 } = ( { 2.7 \times 10 ^ {-15} \over 27 } ) ^ {1/4 } = 1 \times 10 ^ {-4} M $ $ \therefore 2 \times 10 ^ {-4} \gt 1 \times 10 ^ {-4} \gt 2 \times 10 ^ {-5 } $ $ \therefore MX \gt M_3 X \gt MX_2 $
When $ H ^ + $ ion concentration of a solution increases
The pH of a solution is a measure of the concentration of hydrogen ions ($H^+$). When the concentration of $H^+$ ions increases, the pH value decreases because pH is the negative logarithm of the $H^+$ ion concentration (pH = -log[H+]).
The aqueous solution of $ HCOO Na, C_6H_5 NH_3Cl, and KCN $ are respectively
HCOONa is a Salt of weak acid (HCOOH) and Strong base (NaOH) So it is basic. $ C_6H_5NH_3Cl is a Salt of weak base (C_6H_5 NH_2 )$ and strong acid (HCl) so it is acidic. KCN is a Salt of Strong base (KOH) and weak acid (HCN) so it is basic.
$ K_{SP} of AgIO_3 is 1 \times 10 ^ {-8} $ at a given temperature what is the mass of $ AgIO_3 $ in 100 ml of its saturated solution ?
$ AgIO_{3(g)} \rightleftharpoons Ag ^ + _{(aq)} + IO ^- _{3(aq)} $ $ K_{SP} = S^2 $ $ \therefore S = { K_{SP} ) ^ {1/2} = (1.0 \times 10 ^ {-8} ) ^ {1/2} = 1 \times 10 ^ {-4} mol /lit $ $ \therefore S = 1 \times 10 ^ {-4} \times 283 = 283 \times 10 ^ {-4} gm/ lit $ $ 1000 ml contains 283 \times 10 ^ {-4} gm of AgIO_3 $ $ 100 ml contains 28.3 \times 10 ^{-4} gm of AgIO_3 $
PH of a solution containing 50 mg of sodium hydroxide in $10 dm^3 $ of the solution is
Molar concentration of $ NaOH = { 50 \times 10 ^ {-3} gm \over 40 gm mol ^ {-1} \times dm^3 } = 1.25 \times 10^ {-4} M $ $ P ^{OH} = - log ( 1.25 \times 10 ^ {-4} ) $ = -0.0969 + 4.0 = 3.9031 $ \therefore P ^ H = 14 - 3.9031 = 10. 0969 $
Which one of the following has the lowest $ P^H $ value ?
$ 0.1 M HCl means [H^+] = 10 ^ {-1} \therefore P ^H = 1 $
$ 0.1M KOH means [OH^-] = 10 ^ {-1} P^ {OH} =1 \therefore P^H = 13 $
$ 0.01 M HCl means [H ^+] = 10 ^ {-2} \therefore P ^ H = 2 $
$ 0.01M KOH means [OH^-] = 10 ^ {-2} \therefore P ^ {OH} = 2 \therefore P ^ H = 12 $
Equal volumes of three basic solutions of 11,12 and13 are mixed in a vessel. what will be the $ H^+ $ ion concentration in the mixture ?
$ [H ^ + ] ion concentrations are 10 ^{-11}, 10 ^{-12} and 10 ^{-13} on mixing equal volumes, [H^+] $ in final solution $ = { 10 ^ {-11} \times 10 ^ {-12} \times 10 ^ {-13} \over 3 } = { 10 \times 10 ^ {-12} + 1 \times 10 ^ {-12} + 0.1 \times 10 ^ {-12} \over 3 } $ $ = 11.1 \times 10 ^ {-12 } \over 3 } = 3.7 \times 10 ^ {-12} M $
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