KOH and K2O relationship

hey

Can someone help me understand a problem that I am dealing with in greenhouse/fertilizer studies?
A person working in the industry supplied a pH up with the following information:

"We take our 45% KOH which is 37% K2O and dilute down to 10% KOH 8%K2O".

Can anyone help me to understand how I can have a solution that is both 10%KOH and 8%K20.  Does this make sense?

Thanks in Advance

Yes it makes sense. Say you have 100g of solution. It will contain 10 g or KOH and 8 g of KO2

Thanks Seanfisk.

Is there any way you can help me to understand why this may be in solution in this way?
Thanks

Hi 

45%KOH when diluted to 10% it means the volume is changed from 100mL to 450mL 

use M1V1=M2V2

here M1 can be taken as 45% and M2 can be taken as 10%,V1 as 100 % 

 

In the same solution 37% K2O  is present 

again use M1V1 = M2V2 

M1 as 37% ,V1 as 100mL 

M2???

V2= 450mL 

check are you getting 8%(approx)

It's probably in solution that way to balance the pH. It all you have is the KOH that's a strong base, the KO2 will bring extra potassium to the party, while moderating the pH a bit, since KO2 is slightly acidic. Since this is a fertilizer formula, it's preferable to have a slightly alkaline ( basic) solution. The mixture of the two potassium sources will also give it some buffering capacity, so it will resist pH changes too.