00:07
In this problem we have to calculate molarity of solution and ph and poh of the given solution.
00:13
In part a, 25 milligram of h2 -so4 is present in 500 ml of aqua solution.
00:26
Therefore, mass of h2 -s -o -4 denoted by small w is equal to 25 .0 milligram which is equal to 25 .0 milligram which is equal to 25 into 10.
00:45
To the power minus 3 gram.
00:51
Volume of aqua solution denoted by capital v is equal to 500 ml which is equal to 0 .5 liter.
01:04
Molecular weight of h2so4 is equal to 98 gram per mole.
01:18
Therefore, molarity of h2 s of 4 is equal to moles of solute which is h2s4 divide by volume of solution in liter and moles of h2s4 is equal to mass of h2s4 divide by molecular weight of h2sf4 into volume of solution so this is equal to 25 into 10 dashed power minus 3 gram divide by 98 gram per more into 0 .5 liter.
02:15
With the help of calculator, we get this value is equal to 5 .1 into 10 .3 power minus 4 molar.
02:29
In part b, we have to calculate molarity of hcl.
02:41
Which is equal to mass of hcl in gram, divide by molecular weight of hcl into volume of solution in liter.
03:05
So this is equal to mass of hl is 60mg, which is equal to 60 .0 into 10 to the power minus 3 gram.
03:16
Molecular weight of hcl is 36 .5.
03:21
Gram per mole into volume of solution is 500 ml which is equal to 0 .50 liter so this is equal to 3 .3 into 10 to the power minus 3 molar.
03:48
In part c we have to calculate ph of both the solution...