25 drops of acid is required to neutralize the 50.0 ml of 0.010m of NaOH in the experiment.
The equation of the reaction is;
NaOH(aq) + HCl(aq) ---------> NaCl(aq) + H2O(l)
We can use the titration formula;
CAVA/CBVB = NA/NB
CA= concentration of acid
VA = volume of acid
CB = concentration of base
VB = volume of base
NA = number of moles of acid
NB = number of moles of base
CB = 0.010 M
VB = 50.0 ml
CA = 0.50 M
VA = ?
NA = 1
NB = 1
Substituting values;
CAVANB = CBVBNA
VA = 0.010 × 50.0 × 1/ 0.50 × 1
VA = 1 ml
Since the total volume of acid used is 1 ml and each drop contains 0.040 ml
The number of drops required is 1ml/0.040 ml = 25 drops
Learn more: brainly.com/question/1527403
KOH is a strong base, so [OH-] = 0.10 M = 1.0 x 10^-1 M
[H+][OH-] = Kw
Kw = 1.0 x 10^-14
[H+] = 1.0 x 10^-14 / 1.0 x 10^-1 = 1.0 x 10^-13 M
Answer:
C
Explanation:
I just think it is the right answer :) Sorry if I am wrong..
This picture represents electrons
Answer:
D.) 1.71 M NaCl
Explanation:
Molarity equation: M= n/v
n= moles of solute
v=liters of solution
NaCl= 58.443 g/mol
30g NaCl / 58.443g/mol = 0.5133(this is n)
0.5133 mols/0.300 L=1.71115674 M