Answer: to calculate pH use -log[H+] or - log[OH-]..the solution is basic as the “NaOH” is attached to a hydroxide.Since we need to find the pH (per hydrogen) and not the pOH( per hydroxide) we need to find the pOH of the substance first then we subtract that by 14 so we can arrive at the pH of the substance.
Explanation: So -log( 1 x 10^(-5)) = 5 which is the pOH.Now we subtract that by 14 which gives us -9 and now you’d multiply that by -1 bcuz we can’t have a negative so the pH of the substance is 9
Answer:
HClO4(aq) + H2O(l) ⇄ H3O+(aq) + ClO4-(aq)
Explanation:
When an acid dissolves in water, it will produce hydronium ion, H3O+ as the only positive ion.
perchloric acid, HClO4, being an acid will react with water, H2O to produce hydronium ion, H3O+ as shown below:
HClO4(aq) + H2O(l) ⇄ H3O+(aq) + ClO4-(aq)
The Molarity of a solution = number of moles / volume.
Volume = 244ml = 0.244L
So it follows that number of moles = Molarity * volume
Number of moles = 0.135 * 0.244 = 0.03945.
Hence the number of moles = 0.03945
Answer:
pH=11.
Explanation:
Hello!
In this case, since the data is not given, it is possible to use a similar problem like:
"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"
Thus, for the reaction:
Tt is possible to compute the remaining moles of ethylamine via the following subtraction:
Thus, the concentration of ethylamine in solution is:
![[ethylamine]=\frac{0.0816mol}{0.1850L+0.1144L}=0.2725M](https://tex.z-dn.net/?f=%5Bethylamine%5D%3D%5Cfrac%7B0.0816mol%7D%7B0.1850L%2B0.1144L%7D%3D0.2725M)
Now, we can also infer that some salt is formed, and has the following concentration:
![[salt]=\frac{0.0549mol}{0.1850L+0.1144L}=0.1834M](https://tex.z-dn.net/?f=%5Bsalt%5D%3D%5Cfrac%7B0.0549mol%7D%7B0.1850L%2B0.1144L%7D%3D0.1834M)
Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:
![pOH=pKb+log(\frac{[salt]}{[base]} )\\\\pOH=3.19+log(\frac{0.1834M}{0.2725M})\\\\pOH=3.0](https://tex.z-dn.net/?f=pOH%3DpKb%2Blog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D%20%29%5C%5C%5C%5CpOH%3D3.19%2Blog%28%5Cfrac%7B0.1834M%7D%7B0.2725M%7D%29%5C%5C%5C%5CpOH%3D3.0)
Finally, the pH turns out to be:
NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.
Best regards!
Answer is: halogens ans noble gases.
The nonmetals are divided into two categories: reactive nonmetals and noble gases.
Halogen elements are in group 17: fluorine (F), chlorine (Cl), bromine (Br) and iodine (I). They are very reactive and easily form many compounds.
Noble gases are in group 18: helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe) and radon (Rn). They have very low chemical reactivity.
