The HCL concentration is 1.2M or 1.2N
A 25.0 mL sample of HCl was titrated to the endpoint with 15.0 mL of 2.0 M NaOH. What was the molarity of HCl?
12
(25.00ml HBr)( Molarity ) = ( 18.80ml NaOH )( 0.150 M ) Molar concetration of HBr = 0.108 M
3.00 M
Yes, you could possibly have errors in your data about the molarity of a vinegar sample during a titration experiment if there was an error made during weighing.
during a titration when a titrant completely furnished the sample then this is the end point of titration.
1.0 Moles
(25.00ml HBr)( Molarity ) = ( 18.80ml NaOH )( 0.150 M ) Molar concetration of HBr = 0.108 M
3.00 M
The molarity is 6.
Yes, you could possibly have errors in your data about the molarity of a vinegar sample during a titration experiment if there was an error made during weighing.
1.3g
during a titration when a titrant completely furnished the sample then this is the end point of titration.
1.0 Moles
The substance to be analysed normally of unknown quantity is called as analyte. it is called as sample. In titration of analytes normally we take it according to the Normality of the titrant taken to find the quantity of analyte. Sample Size = Titer Value*Normality*Molecularr(or)equivalent weight/ purity/10. This formula is expressed in terms of %.
To find the unknown concentration of a sample by using a reagent with a known concentration. ( IE; molarity )
In an acid-base titration problem, the formula to use is: MaVa = MbVb, where the molarity of the acid times its volume equals the molarity of the base times its volume.Here, we have:Ma(10.00mL) = (0.135M)(31.25mL)Solving for Ma = 4.22M.mL / 10.00mL = 0.422M(Note: This is only valid for monoprotic acid with monoprotic bases only, as in this case. If it were titrated with 0.135M carbonate (CO32-) the findings need to be doubled.)
Molarity = moles of solute(CuSO4)/volume of solution(Liters) 0.967 grams CuSO4 (1 mole CuSO4/159.62 grams) = 0.00606 moles CuSO4 Molarity = 0.00606 moles/0.020 liters = 0.303 Molarity
1.00x10^-7