a student add the same acidic indicator to 1M solutions of carbonic acid and hydrochloric acid:?
397908
The different acids have a different pH so turn a differetn colour.\nThe concentrations are different. The HCl is more concentrated as it will fully dissociate.
892272
It tells you when you're at the end point, and the end point is where ever you want it to be, hopefully past but not too far from the equivalence point. \n\nThis isn't always used for pH. Let's say you wanted to know the concentration Sodium in a Na2C2O4 solution. You could use standardized KMnO4 (which is purple) and do a redox reaction which would make CO2 and a bunch of cations (K, Mn, Na). This would be a redox titration, which is different from acid base. This redox titration is actually pretty interesting because a reagent is the indicator, so the end point is when the solution changes to the color of the reactant (this particular reaction is also self catalyzing, which is kinda neat as well).
724819
To determine the concentration of a reagent.
741771
0.5M = 500mM\nv x c=v'xc'\nv x 500 = 50 x 1\n0.1mL of the 0.5M solution diluted to 50mL will be 1mM
43703
How about the red fountain experiment with acids and litmus solutions? mail me at pr4kane@yahoo.com for furhter details on it. u can try electroplating,bleaching by chlorine/so2,the brown ring test with sulphuric acid etc.
1117999
First make a chemical equation, balance it, and find a mole ratio.  Use that ratio and the above information to solve the problem.
696025
b)carboxylic acid and an amine
568277
Yeah, it can be done with just the line.  \nFirst, locate the end point on the line.  It's exactly the midpoint of the sharp pH rise or drop.  Next, find exactly half the end point, where the moles of salt equal the moles of whatever you're adding.  At that point, pKa = pH.
416321
There is no such thing as a concentrated solution of the carbonic acid, as only a small amount of carbon dioxide will dissolve in water. Also, they are the same color, colorless in the pure form.
320845
Having equal amounts of positive and negative charges.\n\nOr\n\nA mixture of equal acid and base with pH=7.00000
295875
for each one you are given a ratio i am assuming you are combining them both, make sure you write out a balanced equation for each one to make sure, some may not be a 1 to 1 ratio, next Molarity=mol/L therfore 1M = 1 mol per liter, use your periodic table to figure out the molecular weight for each one, each one of these is a strong acid, strong base reaction, therefore they will create water as a product and a salt, in the first one you are given 1.0 mol of HCl which is reacting w/ NaOH 1 to 1, therefore you only need to convert to # of moles of NaOH i am assuming you can do this hopefully this helps
968497
The initial concentration of the acid is C=mole/V=1/2=0.5 mole/L\nLet's assume that the acid is HA\n\n.. .. .. .. .. .. .. HA <=> H(+) + A(-)\nInitial .. .. .. .. 0.5\nDissociate .. .x\nProduce .. .. .. .. .. .. .. x .. .. .. x\nAt Equil. .. . 0.5-x .. .. . x .. .. .. x\n\nKa=[H+][A-]/[HA] =x^2/(0.5-x)\n\nbut x=[H+]=3.4*10^-2 so\n\nKa= ((3.4*10^-2)^2) / (0.5-3.4*10^-2) =2.48*10^-3
554057
Carbonic acid formed by the CO2 in the soda.
423114
Adding sulphuric acid and sodium hydroxide respectively??
851560
solution...Bicarbonate and sodium. :)
438197
This is sulfuric acid and is just as dangerous as hydrochloric acid. However, since you only have one reactant and there is no reaction, you do not have a chemical equation.
186466
A quantity of 22% acid solution is being diluted with some 8% acid solution. The final solution must be 12% acid & contain 101 mL. How much of each solution must be added to accomplish this?- write and solve a system of equations that describes this.\n\nx = amount of the 22% soln\ny = amount of the 8% soln\nx + y = amount of the 12% soln = 101 ml\n\n.22x + .08y = (x + y)*(.12)\n.22x + .08y = .12x + .12y\n.10x - .04y = 0\n\nx + y = 101\nx = 101 - y\n\n.10(101 - y) - .04y = 0\n10.1 - .10y - .04y = 0\n10.1 - .14y = 0\n10.1 = .14y\n(10.1)/(.14) = y\n72.143 ml = y\n\nx + y = 101 ml\nx = 101 - y\nx = 101 - 72.143\nx = 28.857 ml