00:01
The question says to calculate the relative error in the determination of k plus in a solution.
00:20
And that has k plus concentration of 2 .55 multiply by 10 power minus 3 molarity.
00:26
And here the na positive concentration is also given for all the parts.
00:30
For the first part, the na positive concentration is 2 multiply by 10 power minus 2 modality.
00:35
For the second part, it is 2 multiply by 10 power minus 3 molarity.
00:40
And for the third part, it is 2 multiply by 10 power minus 4 molarity.
00:46
So let's see the explanation for it.
00:51
First of all, if we see that selectively coefficient, which is k, k positive, n a positive, this is equal to 0 .052.
01:03
And we know that k, k, k positive, na positive, this will be equal to k, k, k, and a positive, this will be equal to k n a positive divided by k k positive so from here what i can write i can write k positive will be equal to 1 this is equation first and k n a positive this will be equal to 0 .052 this will be equation second then signal of sample s this will be equal to k k positive plus multi it will be multiplied by k k k plus k plus k n a positive multiply by c n a positive and what is given here c k positive is equal to 2 .55 multiply by 10 power minus 3 molarity if we see for the a part c na positive is equal to multiply by 10 power minus 2 molarity so s sam will be equal to 1 multiply by 2 x xx by 10 power minus 3 plus 0 .052 multiply by 2 10 power minus 2.
02:24
So s samp will be equal to 2 .5 10 power minus 3 plus 1 .04 10 power minus 3.
02:34
From here we will get the value of s sample this will be equal to 3 .54 multiply by 10 power minus 3 molarity.
02:43
So the relative error what will be the relative error? the relative error is represented by s -samp minus ck positive divided by ck positive multiply by 100.
03:04
So we will get its value.
03:07
This will be equal to 3 .54 minus 2 .5 multiplied by 10 power minus 3, multiplied by 100, divided by 2 .5 multiply by 10 power minus 3...