The osmotic pressure of 0.05 M monobasic acid having poh= 12 at 25 degree celcius is

We know,pH=pOH +14=14-12=2pH  = 2pH = - Log H3O+H3O+ = 1 ×10-2For monobasic acid, C = concentration of AcidDegree of ionisation, α, =H3O+C = 1 ×10-20.05 =1 ×10-25×10-2 =0.2α =i -1m-1  0.2 =i-12-1=>i=0.2+1=1.2i =1.2π =i CRT = 1.2 ×0.05 RT = 0.06RT (atm)[R=0.082 , T=25+273=298K]

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