How can you ready yourself a barrier provider off pH9

O (ten -7 M) is minimal in comparison to the H

Question 9. 1. You are provided with 0.1 M NH_cuatroOH solution and ammonium chloride crystals. (Given: pKb for NH₄OH is 4.7 at 25°C)

2. What volume of 0.six M sodium formate solution is required to prepare a buffer solution of pH 4.0 by mixing it with 100 ml of 0.8 M formic acid. (Given: pK_a for formic acid is step three.75.) Answer: 1.

We understand that pH + pOH = 14 9 + pOH = 14 = pOH = fourteen – 9 = 5

[NH₄Cl] = 0.1 M x 1.995 = 0. 1995 M escort girl Huntsville =0.2 M Amount of NH₄CI required to prepare 1 litre 0.2 M solution = Strength of NH₄CI x molar mass of NH₄CI = 0.2 x 535 = g g ammonium chloride is dissolved in water and the solution is made up to one litre to get 0.2 M solution. On mixing equal volume of the given NH₄OH solution and the prepared NH₄CI solution will give a buffer solution with required pH value (pH = 9).

[Sodium formate] = number of moles of HCOONa = 0.6 x V x 10 -3 [formic acid] = number of moles of HCOOH = 0.8 x 100 x 10 -3 [formic acid] = number of moles of HCOOH = 0.8 x 100 x 10 -3 = 80 x 10 -3 4 = 3.75 + log \(\frac < 0.6V>< 80>\) 0.25 = log \(\frac < 0.6V>< 80>\) antilog of 0.25 = \(\frac < 0.6V>< 80>\) 0.6V = 1.778 x 80 = 1.78 x 80 = 142.4 V = \(\frac < 142.4>< 0.6>\) = mL

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3. pH of 0.05M sodium carbonate solution pK_a for HCO₃ – is .

Question 1. Identify the Lewis acid and the Lewis base in the following reactions. Cr 3+ + 6H₂O > [Cr(H₂O)₆] 3+ In the hydration of ion, each of six water molecules donates a pair of electron to Cr 3+ to form the hydrated cation, Hexa aqua chromium (III) ion, thus, the lewis acid is Cr and the Lewis base H₂O.

Question 2. Calculate the concentration of OH – in a fruit juice which contains 2 x 10 M, H₃O + ion. Identify the nature of the solution. Answer: Given that H₃O + = 2 x 10 -3 M K_w = [H₃O + ] [OH – ]

O + ] out of ionisation from H

2 x 10 -3 >> 0.5 x 10 -11 i.e., [H₃O + ] >> [OH – ], hence the juice is acidic in nature

H₃O from the auto ionisation of H₂₃O from 10 -3 M HCI. Hence [H₃O + ] = 0.001 mol dm -3 pH = – log₁₀ [H₃O + ] = – log₁₀ (0.001) = – log₁₀ (10 -3 ) = 3

Question 4. Calculate pH of 10 -7 M HCI Answer: If we do not consider [H₃O + ] from the ionisation of H₂O, then [H₃O + ] = [HCl] = 10 7 M i.e., pH =7, which is a pH of a neutral solution. We know that HCI solution is acidic whatever may be the concentration of HCI i.e, the pH value should be less than 7. In this case the concentration of the acid is very low (10 -7 M). Hence, the H₃O + (10 -7 M) formed due to the auto ionisation of water cannot be neglected. So, in this case we should consider [H₃₂O Answer: [H₃O + ] = 10 -7 (from HCl) + 10 -7 (from water) = 10 -7 (1+1) = 2 x 10 pH = – log₁₀ [H₃O] = – log₁₀(2 x 107) = – [log2 + log 10 -7 ] = – log 2 – ( – 7). 1og₁₀ = 7 – log 2 = 7 – 0.3010 = 6.6990 = 6.70