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The ionzation constant of formic acid is $ 7.8 \times 10 ^{-4 }. Calculate ratio of sodium formate & formic acid in a buffer of P^H 4.25 $
$ pka = -log ( 1.8 \times 10 ^ {-4} ) = 3.74 $ $ log { [salt ] \over [acid] } = P^H - pka = 4.25 - 3.74 = 0.51 $ $ \therefore { [salt] \over [acid] } = antilog of 0.51 = 3.24 $
The ionization constants of HF is $6.8 \times 10 ^{-4} $ . Calculate ionization constant of corresponding conjugate base.
$ K_b = { kw/kc} = 10 ^ {-14} \over 6.8 \times 10 ^ {-4} = 1.47 \times 10 ^ {-11} $
The ionization constant of formic acid is $1.8 \times 10 ^{-4} around what P^H $ will its mixture with sodium formate give buffer solution of highest capacity
Buffer Solution of highest capacity is formed at which $ P^H = pka = - log(1.8 \times 10 ^{-4}) = 3.74 $
What is PH of our blood ? why does it remains constant inspite the variety of the foods and spices we eat ?
The pH of blood is approximately 7.4. It remains constant because blood acts as a buffer solution. A buffer solution can resist changes in pH when small amounts of acid or base are added. This buffering capacity is crucial for maintaining the proper functioning of biological systems.
The precipitate of $CaF_2 (K_{SP} = 1.7 \times 10 ^{-10} ) $ is obtained when equal volumes of the following are mixed.
Ionic Product of $ CaF_2$ in (i) $ IP = [Ca^{2+}] [ F ^ -] ^ 2 = 10 ^ {-12} \lt ksp $ (ii) $ IP = 10 ^ {-2} \times (10 ^ {-3} ) ^ 2 = 10 ^ {-8} \gt ksp \therefore ppt .obtain $ (iii) $ IP = 10 ^ {-5 } \times ( 10 ^ {-3} ) ^2 = 10 ^ {-11} \lt ksp $ (iv) $ IP = 10 ^ {-3} \times (10 ^ {-5} ) ^2 = 10 ^ {-13} \lt ksp $
The correct order of increasing $ [H_3O^+] $ in the following aqueous solution is.
$ H_2 S =weak acid H_2 SO_4 =strong acid $ $ NaCl = neutral NaNO_2 = basic $ $ Hence [ H_3 O ^ + ] will be in the order of NaNO_2 \lt NaCl \lt H_2 S \lt H_2 SO_4 $
What is the % hydrolysis of NaCN in N/80 solution when dissociation constant for HCN is $ 1.3 \times 10 ^{-9} \& K_w = 1 \times 10^{-14} $
NaCN is Salt of weak acid (HCN) and Strong base (NaOH) Hence $ h = \sqrt { kw \over k_a C } = \sqrt { 10 ^ {-14 } \over (1.3 \times 10 ^ {-9} )\times {1 \over 80 } } = 2.48 \times 10 ^ {-2} $ $ \therefore Percentage Hydrolysis = ( 2.48 \times 10 ^ {-2} ) 100 = 2.48 $
The $ K_{SP} of AgCl is 4.0 \times 10 ^ {-10} at 298 k.solubility of AgCl is 0.04 M CaCl_2 $ will be
If x is Solubility ofAgCl in 0.04 M $ CaCl_2 $ , then
$ [Ag^+] = x mol L ^ {-1} $
$ [Cl^1] = (0.04 \times 2 ) + x = 0.08 M $
$ 0.08 (x) = 4 \times 10 ^ {-10} $
$ x = 5.0 \times 10 ^ {-9} M $
How much sodium acetate should be added to 0.1 M solution $ CH_3COOH to give a solution of P^H 5.5 (pKa of CH_3CooH = 4.5 ) $
$ P^H = pka + { log [CH_3COONa] \over 1 [CH_3COOH] $ $ 5.5 = 4.5 + log { [CH_3COONa] \over 0.7 } $ $ 4.5 + log [CH_3COONa ] + 1 $ $ \therefore log [CH_3COONa] = 0 $ $ \therefore [CH_3COONa] = 7 M $
The $ P^H of solution obtained by mixing 50ml 0.4 N HCl \& 50ml 0.2 N NaOH $ is
$ 50 ml of 0.4 NHCl = { 0.4 \over 1000 } \times 50 = 0.02 g eq $ $ 50 ml of 0.2 N NaOH = { 0.2 \over 1000 } \times 50 = 0.01 g eq $ 0.01 g eq of NaOH will Neutalilise 0.01 g eq of HCl $ \therefore HCl left unneutralised = 0.01 g eq vol of Sol. =50+50 =100ml$ $ \therefore [HCl] = { 0.01 \over 100} \times 1000 = 0.1 N $ $ or [ H^+] = 0.1 M $ $ \therefore P ^ H = log (0.1) = 1.0 $
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