Practice free NEET NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
With increase in temperature, ionic product of water
With increase in temperature, ionic product increases. because self ionisation of is endothermic process
EDTA is a/an
EDTA is Arrhenius acid as it can give $ H ^ + $ ions in aqueous solution, bronsted base. as it can accept protons and lewis base because N and O in it can donate lone paris of electrons.
The units of lonic product of water (kw) are :
$ K_w = [ H^+ ] [ OH^ - ] $ $ = mol L ^ {-1} \times mol L ^ {-1} = mol ^ 2 L ^ { -2 } $
Which of the following weakest ?
Smaller the Kb value, weaker is the base.
On adding ammonia to water
On adding$ NH_3 to water, [OH ^-] will increase, Kw = [H_3O^+][OH ^ - ] is constant.$ Therefore $ [H_3O ^ + ] $ will decrease
The $ P^H$ of a buffer containing equalmolar concentration of a weak base and its chlorides is $ (K for weak base = 2 \times 10 ^ {-5} )(log 2 = 0.3010)$
$ \therefore P^{OH} = P^Kb + log {[Salt] \over [Base] } [Salt] = [ Base ] $ $ = - log(2 \times 10 ^ {-5} ) + log1 $ $ = - 0.3010 + 5.000 + 0 = 4.6990 $ $ P^H = 14 - p ^ {OH} = 14 - 4.6990 = 9.3010 $
Solution of 0.1 N $NH_4OH and 0.1N NH_4Cl has P^H 9.25.Then P_{K_b}$ is
$ P^{OH} = p^Kb + log { [Salt] \over [Base]} $
$ P^{OH} =14 - P^H = 4.75 $
$ 4.75 = p^{kb} + log1 $
$ pk_b = 4.75 ( log 1 = 0 )
The solubility of product barium sulphate is $ 1.5 \times 10 ^ {-9}$ at $18 ^\circ$ C Its solubility in water at $18 ^\circ C $ is
$ BaSO_{4(s)} \rightleftharpoons Ba^{2+} _{(aq)} + So ^ {2-}_{4(aq)} $ xmol/lit x mol/lit $ K_{SP} = [Ba^{2+} ][ So ^{2-} _ { 4} ] $ $ K_{SP} = X^2 $ $ \therefore x = (ksp) ^ { 1 /2 } = (1.5 \times 10 ^ {-9} ) ^ {1/2} = 3.9 \times 10 ^ { -5} mol L ^ { -1} $
The least soluble compound (salt) of the following is
Hint : As value of $ K_{SP} $ is less , solubility is also less.
What is the value of $ P^H of 0.01 M glycine solution ? For glycine Ka_1 = 4.5 \times 10 ^ {-3} and Ka_2 = 1.7 \times 10 ^ {-10} $ at 298k.
$ glycine (NH_2CH_2COOH)$ is more acidic than basic. overall ionisation constant $ K = K_{a1} \times K_{a2} = 4.5 \times 10 ^ {-3} \times 1.7 \times 10 ^ {-10} $ $ = 7.65 \times 10 ^ {-13} $ $ [H^+] = \sqrt{ K.C } = \sqrt { 7.65 \times 10 ^ {-13} \times 0.01 } $ $ = 0.87 \times 10 ^ { -7} M $ $ P ^ H = -log ( 8.7 \times 10^{-8} ) = 7.0605 $
Ready to ace NEET?
Free access · No credit card required
Yes. You can attempt every NEET question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.
No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.
The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.