Answer:
1.7 mL
Explanation:
<em>A chemist must prepare 550.0 mL of hydrochloric acid solution with a pH of 1.60 at 25 °C. He will do this in three steps: Fill a 550.0 mL volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (8.0 M) stock hydrochloric acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated hydrochloric acid that the chemist must measure out in the second step. Round your answer to 2 significant digits.</em>
Step 1: Calculate [H⁺] in the dilute solution
We will use the following expresion.
pH = -log [H⁺]
[H⁺] = antilog - pH = antilog -1.60 = 0.0251 M
Since HCl is a strong monoprotic acid, the concentration of HCl in the dilute solution is 0.0251 M.
Step 2: Calculate the volume of the concentrated HCl solution
We want to prepare 550.0 mL of a 0.0251 M HCl solution. We can calculate the volume of the 8.0 M solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂/C₁
V₁ = 0.0251 M × 550.0 mL/8.0 M = 1.7 mL
Answer:
pH = 2.46
Explanation:
Hello there!
In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:
![n_{acid}=n_{base}=n_{salt}](https://tex.z-dn.net/?f=n_%7Bacid%7D%3Dn_%7Bbase%7D%3Dn_%7Bsalt%7D)
Whereas the moles of the salt are computed as shown below:
![n_{salt}=0.021L*0.68mol/L=0.01428mol](https://tex.z-dn.net/?f=n_%7Bsalt%7D%3D0.021L%2A0.68mol%2FL%3D0.01428mol)
So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:
![[salt]=0.01428mol/0.0276L=0.517M](https://tex.z-dn.net/?f=%5Bsalt%5D%3D0.01428mol%2F0.0276L%3D0.517M)
Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:
![C_6H_5NH_3^++H_2O\rightleftharpoons C_6H_5NH_2+H_3O^+](https://tex.z-dn.net/?f=C_6H_5NH_3%5E%2B%2BH_2O%5Crightleftharpoons%20%20C_6H_5NH_2%2BH_3O%5E%2B)
Whose equilibrium expression is:
![Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BC_6H_5NH_2%5D%5BH_3O%5E%2B%5D%7D%7BC_6H_5NH_3%5E%2B%7D)
Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:
![2.326x10^{-5}=\frac{x^2}{0.517M}](https://tex.z-dn.net/?f=2.326x10%5E%7B-5%7D%3D%5Cfrac%7Bx%5E2%7D%7B0.517M%7D)
Whereas x is:
![x=\sqrt{0.517*2.326x10^{-5}}\\\\x=3.47x10^-3](https://tex.z-dn.net/?f=x%3D%5Csqrt%7B0.517%2A2.326x10%5E%7B-5%7D%7D%5C%5C%5C%5Cx%3D3.47x10%5E-3)
Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:
![pH=-log(3.47x10^{-3})\\\\pH=2.46](https://tex.z-dn.net/?f=pH%3D-log%283.47x10%5E%7B-3%7D%29%5C%5C%5C%5CpH%3D2.46)
Regards!
Answer:
on https://www.ck12.org/book/peoples-physics-concepts/section/20.1/
Explanation:
it shows for calcium 45 and maybe a possible equation on 3/4th's scroll down?
M C C O R M I C K T E A M ’ S O P E N Q U A N T U M M AT E R I A L S D ATA B A S E
O F F E R S U N L I M I T E D A C C E S S T O
ANALYSES OF NEARLY 300,000 COMPOUNDS
The answer is E.
Chemoautotrophs are organisms that use inorganic <span>compounds as an energy source and CO2 as a carbon source. The organisms live in </span>hostile environments like dark depths of the sea. Inorganic energy sources that they use are mostly hydrogen sulfide, elemental sulfur, ferrous iron, molecular hydrogen, and ammonia.