## How can you prepare a barrier service regarding pH9

How can you prepare a barrier service regarding pH9
O (ten -7 M) are minimal when compared to the H

Question 9. 1. You are provided with 0.1 M NHcuatroOH 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: pKa for formic acid is step step three.75.) Answer: 1.

We all know that pH + pOH = fourteen nine + pOH = 14 = pOH = fourteen – nine = 5

[NH4Cl] = 0.1 M x 1.995 = 0. 1995 M =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 pKa 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 Kw = [H3O + ] escort girl Riverside [OH – ]

## O + ] away from ionisation off 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

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