Answer:
3.336.
Explanation:
<em>Herein, the no. of millimoles of the acid (HCOOH) is more than that of the base (NaOH).</em>
<em />
So, <em>concentration of excess acid = [(NV)acid - (NV)base]/V total</em> = [(30.0 mL)(0.1 M) - (29.3 mL)(0.1 M)]/(59.3 mL) = <em>1.18 x 10⁻³ M.</em>
<em></em>
<em> For weak acids; [H⁺] = √Ka.C</em> = √(1.8 x 10⁻⁴)(1.18 x 10⁻³ M) = <em>4.61 x 10⁻⁴ M.</em>
∵ pH = - log[H⁺].
<em>∴ pH = - log(4.61 x 10⁻⁴) = 3.336.</em>
Answer:
True
Explanation:
The molecule CH20 contains two single bonds and one double bond.
Answer:
pH of buffer =4.75
Explanation:
The pH of buffer solution is calculated using Henderson Hassalbalch's equation:
![pH=pKa+log[\frac{[salt]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%5B%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
Given:
pKa = 3.75
concentration of acid = concentration of formic acid = 1 M
concentration of salt = concentration of sodium formate = 10 M
![pH=3.75+log[\frac{10}{1}]=3.75+1=4.75](https://tex.z-dn.net/?f=pH%3D3.75%2Blog%5B%5Cfrac%7B10%7D%7B1%7D%5D%3D3.75%2B1%3D4.75)
pH of buffer =4.75
Answer:
The correct answer is 8.79 × 10⁻² M.
Explanation:
Based on the given information, the mass of NaI given is 4.11 grams. The molecular mass of NaI is 149.89 gram per mole. The moles of NaI can be determined by using the formula,
No. of moles of NaI = Weight of NaI/ Molecular mass
= 4.11 / 149.89
= 0.027420
The vol. of the solution given is 312 ml or 0.312 L
The molarity can be determined by using the formula,
Molarity = No. of moles/ Volume of the solution in L
= 0.027420/0.312
= 0.0879 M or 8.79 × 10⁻² M
