Concentration of HCl when it is mixed with NaOH?

Question

What is the concentration of HCl after 10.00 mL of 1.00 M HCl is mixed with 10.00 mL of .500 M NaOH?

I think I need to find the limiting reactant...

.01 L x (1.00 mol HCl/ L) x (1 mol H2O/ 1mol HCL) = .0100 mol H20

.01 L x (0.500 mol NaOH/ L) x (1mol H2O/ 1mol NaOH) =
.005 mole H2O.

So, the LR is NaOH.

What is left over is HCl, but I dont what to do next, or if I am doing right at all.

Thanks for your help.

Answer 1

You need to find how many moles of acid and base you originally had. In a way tou have already done it...
mole HCl = C*V= 1*0.01=0.01
mole NaOH =C*V= 0.5*0.01=0.005

The two will react;

NaOH + HCl -> NaCl + H2O

It is 1:1 stoichiometry, so the number of moles HCl reacting is the same as the number of NaOH. Thus the number of HCl moles remaining is molesHCl = molesHCl initial- moles HClreacted =moles HCl initial -moles NaOH =0.01-0.005= 0.005

Your new volume is the volume of the acid + the volume of the base, so V=0.01+0.01= 0.02L

C=mole/V= 0.005/0.02 = 0.250 M

Answer 2

Use a pair of drops of phenothalien indicator on your NaOH decision. Titrate with a known concentration of HCl (in all probability a million M HCl or much less molar concentrations are used.). Titration is including the HCl technique on your NaOH answer drop via ability of drop from a buret. The burette ought to have volumetric markings alongside the side. be conscious the commencing up quantity of HCl. in specific cases gently swirl your NaOH decision. As immediately as you start to make certain a colour commerce, stop your titration after each and every drop and swirl. as quickly as the colour exchange is everlasting, the titration is accomplished. record the recent quantity of HCl. Calculate the quantity of HCl used to neutralize the NaOH. With that quantity, the extensive-unfold molarity of the HCl, and the traditional quantity of NaOH... you need to calcualte the molarity of the hollow NaOH decision. MaVa=MbVb the place M is molarity, V is quantity, a is acid and b is base.

Answer 3

no here you don't need to find a limiting reactent.but i think you should find the concentration of HCl after the mixing so:

n before mixing =n after mixing
CxV = CxV
1.00x10.00x10^-3=Cx(10+10)
therefore C=0.01/20x10^-3=0.5 mol/l