Hi: A 500 ml container has 16.6 oz in it. I already know that if 1oz of 35% is added to 11 oz of distilled that a 3% results but you indicate that adding less than an ounce to 500 ml would give me 3%. Adding less than an ounce to 16.6 oz would give an even weaker solution wouldn't it?
Ian ----- Original Message ----- From: Marshall Dudley To: [email protected] Sent: Tuesday, December 03, 2002 3:37 PM Subject: Re: CS>Titration Math Ian Roe wrote: Hi: I'm not the greatest when it comes to titration mathematics. Can someone here show me the formula for dilution. I knew how to do this once but I just can't remember now. Problem: wish to put an unknown volume of 35% H2O2 into a 250 ml and a 500 ml container and fill with distilled water to obtain 3%. That is pretty easy. You want 500 ml of 3% H2O2, so you want the container to contain .03*500 = 15 ml of H2O2 with the remainder water. Since the H2O2 is 35% to start with you will have to start out with 15/.35 ml of the 35% H2O2 or just under 43 ml. So put just under 43 ml of your 35% H2O2 in the 500 ml container, and fill with water to 500 total volume. Now that will be a volume percentage. For a weight (mass) percentage, then you would have to work with weights, but I think the 35% H2O2 is close enough to the mass of water to not worry about it. Marshall Equipment: 35% H2O2, 250 and 500 ml containers and a 30 ml measuring cup divided off in 5 ml lines, drams at the 1/4 oz markers, 1/4 oz markers - and of course distilled water. Thanking you in advance. IanRoe
