00:01
So here the question says that how many ml of a 0 .50m solution of naoh would be required to neutralize 250 ml of a 0 .70m h solution? okay.
00:15
In addition to the amount, you must provide the complete and balanced equation of the reaction and also how many ml of 0 .75mh3404 if we require to diatriate 200 ml of 0 .5 .0404 if we require to diatriate 200 ml of 0 .5.
00:30
50m c o h2 to the equivalence points.
00:36
Alright, there's a key question too.
00:38
Alright, so let's get to.
00:40
So we have the n -a -o -h -a -c -l plus h -c -l.
01:02
This goes to nacl plus h2.
01:24
So the number of moves of hcl present becomes equal to 0 .70m multiplied by 250mm of 1000.
02:00
So we have 0 .175 in all.
02:17
Okay, so by street carameterre we have one more hcl.
02:51
Neutralized by one more n -a -o -h 0 .175 mole of h -c -l will be neutralized, neutralized by okay so we have one multiplied by 0 .175 more n -a -o -h so we have 1 .175 more n a oh so that this is equal to 0 .175 no n a oh which so m l h c l so the notion that is required we have 0 .50m is equal to the mole of o h over the volume in m l multiplied by thousand so this gives us volume to be equal to 1 over 0 .50 multiplied by 0 .175 multiplied by 1000 which is equal to 350m .l so 350 ml of 0 .50m .0 m m a .h solution is required...