What volume of 0.0293 M Ba(OH)2 is required to neutralize 25.00 mL of 0.200 M HNO3?
Absolutely, with no offending, the first person's both ways are wrong. There is no way you can use the formula of C1V1=C2v2 with titration process(neautralization) because if you use it, your assuming that both solutions has the same volume and same concentration which is wrong! you only use this equation during dilution of ONE solution, the volume and concentration of it are equal because moles never change in dilution Solution: write the equation first (always write the equation in any chemistry problem to make everything clear). In fact, especially in this problem, you must write down the equation Ba(OH)2(aq)+2HNO3(aq)--->Ba(NO3)2(aq)+... Products in this equation are not necessary in this problem. First you must get the moles of Ba(OH)2 in order to get the volume. To know the moles of Ba(OH)2, the moles of HNO3 first by multiplying volume in L by molarity as follows: 0.025*0.200=0.005 moles Notice that there is A "2" in front of HNO3 in the equation but there is A "1" in front of Ba(OH)2. This simply means that the moles of HNO3 will be DOUBLE the moles of Ba(OH)2 so to get the moles of Ba(OH)2 we will divide 0.005/2=0.0025 moles Now you know the volume of Ba(OH)2... its easy to get its volume Moles=volume(L)*concentration(M) 0.0025=x*0.0293.... doing the simple mass to get the X, which is the volume, 0.0025/0.0293=0.085 L or 85 ML I'm 100% sure of my answer...(I always get A's in tests in chemistry because its my favourite subject) I wrote too much but I just wanted to make sure you understand I hope that helped and good luck!
In order to completely neutralize 20 mL of a solution of HCl 0.1 M, 40 mL of a solution of NaOH must be added. What is the M of the NaOH solution?
No, this working is incorrect. You tripped up on two points.20 milliliters is what decimal fraction of a liter?milli- means one thousandth, so 20 mL = 20/1000 liters20/1000 = 2/100 = 1/50 literDivide one by fifty:50 ) 1 wont go , result 0.? so try50) 10 won’t go result 0.0? so try50)100 goes twice result 0.02So 0.02 L = 20mL Result for VaYou made the same mistake with Vb = 40 mL SHOULD BE = 0.04 LThe second mistake was more important because in this equation the first two scaling mistakes canceled out!You wrote: 1 mol x 0.1 M x 0.20 L / 1 mol x 0.40 L = 8x10-3M1 x 0.1 x 0.2 / 1 x 0.4 = 0.05M NOT 8 millimol !You multiplied (x) instead of dividing ( / )
How many milliliters of water must be added to 50ml of 7.0 M of HCL so that the solution concentration is reduced to 0.35 M?
Molarity= (number of moles of that substance) / (Volume of solution in litres)7= (number of moles of HCl) / 0.05Thus,Number of moles of HCl = 7 x 0.05 = 0.35The number of moles of HCl will remain unchanged.Now, for new solution,0.35= 0.35 / (New volume of solution in litres)Thus,New volume = 1 litreThus, amount of water added = 1- 0.05 litre = 0.95 litre = 950 mlI am pretty sure that's how you do it!
Calculating pH during titration?
100.0 mL of 0.100 M acetic acid is placed in a flask and is titrated with 0.100 M sodium hydroxide. An appropriate indicator is used. Ka for acetic acid is 1.7 s 10^-5. Calculate the pH in the flask at the following points in the titration. a) When no NaOH has been added b) After 25.0 ml of NaOH is added c) After 50.0 ml of NaOH is added d) After 75.0 ml of NaOH is added e) After 100.0 ml of NaOH is added f) After 300.0 ml of NaOH is added I just want to make sure that my math is right... For A- I got a pH of 1 by doing the -log of ((.100 M x .100 L) - (.100 M x 0 L)) // .100 L. I used the same technique to get b) 1.22 c) 1.48 d) 1.85 Is that correct? And then I'm having a hard time with e because with what I'm doing I get -log(0) which of course is undefined.... Help would be appreciated!
A 50.0 mL sample of 0.15M NaOH is added to 50.0 mL of 0.10M Ba(OH)2. What is the molar concentration of OH^-(a?
First find out how many mmol of OH- you will have. (50 ml)(.15 mmol/mL) + (2)(50 mL)(.10 mmol/mL) = 17.5 mmol This is dissolved into 100 mL water, so the concentration would be 17.5 mmol/100 mL = 0.175M --> 0.18M OH- so the answer is c.