Answer:
3.3535 g
Explanation:
Considering:
Or,
Given :
For iron(III) nitrate :
Molarity = 0.404 M
Volume = 52.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 52.0×10⁻³ L
Thus, moles of iron(III) nitrate :
Moles of iron(III) nitrate = 0.021 moles
1 mole of iron(III) nitrate forms 1 mole of iron(III) hydroxide which further forms 1 mole of iron(III) oxide.
Thus, moles of iron(III) oxide formed from 0.021 moles of iron(III) nitrate = 0.021 moles
Also, molar mass of iron(III) oxide = 159.69 g/mol
So, mass of iron(III) oxide = 0.021 moles × 159.69 g/mol = 3.3535 g
The new pH is 7.69.
According to Hendersen Hasselbach equation;
The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.
pH = pKa + log10 ([A–]/[HA])
Here, 100 mL of 0.10 m TRIS buffer pH 8.3
pka = 8.3
0.005 mol of TRIS.
∴ ![8.3 = 8.3 + log \frac{[0.005]}{[0.005]}](https://tex.z-dn.net/?f=8.3%20%3D%208.3%20%2B%20log%20%5Cfrac%7B%5B0.005%5D%7D%7B%5B0.005%5D%7D)
<em> </em>inverse log 0 = ![\frac{[B]}{[A]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D)
![\frac{[B]}{[A]} = 1](https://tex.z-dn.net/?f=%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D%20%3D%201)
Given; 3.0 ml of 1.0 m hcl.
pka = 8.3
0.003 mol of HCL.
![pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69](https://tex.z-dn.net/?f=pH%20%3D%208.3%20%2B%20log%20%5Cfrac%7B%5B0.005-0.003%5D%7D%7B%5B0.005%2B0.003%5D%7D%5C%5CpH%20%3D%208.3%20%2B%20log%20%5Cfrac%7B%5B0.002%5D%7D%7B%5B0.008%5D%7D%5C%5C%5C%5CpH%20%3D%208.3%20%2B%20log%20%7B0.25%7D%5C%5C%5C%5CpH%20%3D%208.3%20%2B%20%28-0.62%29%5C%5CpH%20%3D%207.69)
Therefore, the new pH is 7.69.
Learn more about pH here:
brainly.com/question/24595796
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Answer:
The concentration is 0.036 mg/mL
Explanation:
Concentration = 0.2 mM = 0.2/1000 = 2×10^-4 M = 2×10^-4 mol/L × 180,000 mg/1 mol × 1 L/1000 mL = 0.036 mg/mL
You just have to isolate x, so just divide by WQ on both sides to get: