Question

A 100. 0 ml sample of 0. 10 m nh3 is titrated with 0. 10 m hno3. Determine the ph of the solution after the addition of 100. 0 ml of hno3. The kb of nh3 is 1. 8 × 10-5.

Answer 1

The pH of the solution after the addition of 100. 0 ml of HNO3. The kb of NH3 is 1. 8 × 10-5. So, pH is 10.9 basic .

This neutralization occurs as the acid is added to the base:

NH3(aq)+ HNO3(aq)= NH4NO3(aq)+ H2O(l)

The initial moles of NH3present is given by,

nNH3= c×v= 0.10× 100/1000= 0.01m

The number of moles of

HNO3 added is given by:

nHNO3= c×v= 0.10× 100/1000= 0.001

It is clear from the equation that the acid and base react in a 1:1 molar ratio. So, the no. moles of NH3remaining will be 0.01 - 0.001= 0.009

The total volume is now

100.0+100.0= 200.0x cm3

The concentration of NH3 is given by,

[NH3]= c/v= 0.001/200/1000= 0.05x mol/l .

pOH= 12 (pKb− logb)

where b is the base's concentration.

Given that there is little dissociation, we can roughly compare this to the starting concentration.

pKb= −logKb= −log(1.8×10−5)= 4.744

pOH= 3.056

At 25∘xC, we know that, pH+ pOH=14

pH= 14− 3.056= 10.9

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