You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**kjames****Guest**

HELP!!!

A 90% KCN solution is diluted with water to make a solution of 60% acid. When 3 gallons of water is added to dilute it again, the solution becomes 40% acid.

a. how much water was added to the solution on the first instance?

b. what was the original volume of the 90% acid solution?

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

K = the amount of KCN

w = the amount of water

a = amount of water added

K = 2 / 5

w = 18 / 5

a = 2

You should be able to answer a and b now.

By the way, KCN is the salt of a strong base KOH and a weak acid HCN therefore is it a base not an acid.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

Pages: **1**