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10ml of 0.1 M HCl is titrated with 0.1 M NaOH .when the volume of NaOH added from.the beurette ,from 9.99 ml to 10.01 ml,then ph jump approximately froma) 4 to 10b) 6 to 8c) 6.9 to 7.1d) 1 to 4

10ml of 0.1 M HCl is titrated with 0.1 M NaOH .when the volume of NaOH added from.the beurette ,from 9.99 ml to 10.01 ml,then ph jump approximately froma) 4 to 10b) 6 to 8c) 6.9 to 7.1d) 1 to 4

Grade:11

1 Answers

dip
24 Points
7 years ago
initially there were .01 *0.1=0.001 moles of HCl  or H+(no of moles =Molarity *volume in litre).
When there is 9.99 ml of NaOH in solution ,then solution contains 0.00999*0.1=0.000999 moles of naoh or oh- which neutralise 0.000999 moles of h+ ions.
Therefore no of H+ ions present now=0.000001moles
concentration of h+=10^-6/volume of soln=10^-6/(10+9.99)*10^-3
ph=-log(concentration of h+)=4
similarly ph after 10.01 ml naoh added=10
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