Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
<em />
The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>
Answer:
this doesnt make sence ezxplain the subject
Explanation:
Hello!
First, we need to determine the pKa of the base. It can be found applying the following equation:

Now, we can apply the
Henderson-Hasselbach's equation in the following way:
![pH=pKa+log( \frac{[CH_3NH_2]}{[CH_3NH_3Cl]} )=10,65+log( \frac{0,18M}{0,73M} )=10,04](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%20%5Cfrac%7B%5BCH_3NH_2%5D%7D%7B%5BCH_3NH_3Cl%5D%7D%20%29%3D10%2C65%2Blog%28%20%5Cfrac%7B0%2C18M%7D%7B0%2C73M%7D%20%29%3D10%2C04)
So,
the pH of this buffer solution is 10,04Have a nice day!
Answer:
it has six significant figures
Answer:
16.8 g of AgCl are produced
Explanation:
The reactants are: NaCl and AgNO₃
The products are: AgCl, NaNO₃
Balanced equation: NaCl(aq) + AgNO₃(aq) → NaNO₃(aq) + AgCl(s) ↓
We convert the mass of AgNO₃ to moles → 10 g / 85g/mol = 0.117 moles
Ratio is 1:1, therefore 0.117 moles of nitrate will produce 0.117 moles of AgCl.
According to stoichiormetry.
We convert the moles to mass → 0.117 mol . 143.3g /1mol = 16.8 g