Answer:
Explanation:
Hello,
In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:
![pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1](https://tex.z-dn.net/?f=pH%3D-log%28Ka%29%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5CpH%3D-log%286.3x10%5E%7B-5%7D%29%2Blog%28%5Cfrac%7B0.33M%7D%7B0.42M%7D%20%29%5C%5C%5C%5CpH%3D4.1)
Regards.
Answer:
43.75 ml
Explanation:
Given that the equation of the reaction is;
2HNO3(aq) + Ca(OH)2(aq) ---> Ca(NO3)2(aq) + 2 H20(l)
Concentration of acid CA= 0.05 M
Concentration of base CB = 0.02 M
Volume of acid VA = 35.00ml
Volume of base VB= ???
Number of moles of acid NA= 2
Number of moles of base NB=1
From
CAVA/CBVB= NA/NB
Making VB the subject of the formula;
VB= CAVANB/CBNA
VB= 0.05 × 35 × 1/ 0.02 × 2
VB=1.75 /0.04
VB= 43.75 ml
Gallium sulphate I belive thats it dont count me on that tho
Answer:
Explanation:
In the qualitative analysis of metal salts , we see that in group I , metal chlorides are precipitated out . It is so because their metal chlorides are insoluble in water .
In this group following metal ions are present
Ag+,
Hg₂²⁺
Pb²⁺
1) Astronomers use different measurement systems to discuss distance in the universe C) because they allow the numbers to be more manageable.
The numbers are so huge that astronomers need different units to express them.
2) The measurement system with the largest value is C) the Parsec
.
1 Parsec = 3.26 ly = 2.06 × 10⁵ AU
