00:01
Hello guys, so in this question we have to calculate the ph of the following solutions.
00:05
So the first one is 0 .025 molar of hcl.
00:14
So as the equations are ph is equal to minus of log of h plus and ph plus p .oh is equal to 14.
00:33
So as we have in this question we have, we have already the concentration of h plus so we can find it ph is equal to minus of log of 0 .025 molar which will be equal to our answer that is 1 .6 so this is the answer to the first part of the question next we have been given is ph of we have been given the concentration of hno 3 that is 0 .00 69 molar of hno3 and we have to tell the ph.
01:17
So again we have to write the same formula that will be ph is equal to minus of log of concentration.
01:25
So concentration here is 0 .0069 which will be equal to 2 .16.
01:36
This is the answer.
01:39
Next is we have to find the ph of 0 .054 molar of h2so4 and now when we put this into the equation formula that is equal to minus log of it will be 2 multiplied by 0 .0554.
02:10
So why we have multiplied 2 is because there are 2x plus present in this compound that is h2so4.
02:18
So h2 so4.
02:19
So h2 is 4 will give 2 moles of h plus.
02:27
So it will be equal to 0 .95.
02:33
This is the answer.
02:35
Next we have been given is the concentration of naoh that is 0 .3048 molar of naoh.
02:54
So here in this case we have p -o -h instead of p .s...