<u>Given information:</u>
Concentration of NaF = 0.10 M
Ka of HF = 6.8*10⁻⁴
<u>To determine:</u>
pH of 0.1 M NaF
<u>Explanation:</u>
NaF (aq) ↔ Na+ (aq) + F-(aq)
[Na+] = [F-] = 0.10 M
F- will then react with water in the solution as follows:
F- + H2O ↔ HF + OH-
Kb = [OH-][HF]/[F-]
Kw/Ka = [OH-][HF]/[F-]
At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x
10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x
x = [OH-] = 1.21*10⁻⁶ M
pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92
pH = 14 - pOH = 14-5.92 = 8.08
Ans: (b)
pH of 0.10 M NaF is 8.08
Answer:
0.808 M
Explanation:
Using Raoult's Law
![\frac{P_s}{Pi}= x_i](https://tex.z-dn.net/?f=%5Cfrac%7BP_s%7D%7BPi%7D%3D%20x_i)
where:
= vapor pressure of sea water( solution) = 23.09 mmHg
= vapor pressure of pure water (solute) = 23.76 mmHg
= mole fraction of water
∴
![\frac{23.09}{23.76}= x_i](https://tex.z-dn.net/?f=%5Cfrac%7B23.09%7D%7B23.76%7D%3D%20x_i)
![x_i = 0.9718](https://tex.z-dn.net/?f=x_i%20%3D%200.9718)
![x_i+x_2=1](https://tex.z-dn.net/?f=x_i%2Bx_2%3D1)
![x_2 = 1- x_i](https://tex.z-dn.net/?f=x_2%20%3D%201-%20x_i)
![x_2 = 1- 0.9718](https://tex.z-dn.net/?f=x_2%20%3D%201-%200.9718)
![x_2 = 0.0282](https://tex.z-dn.net/?f=x_2%20%3D%200.0282)
------ equation (1)
------ equation (2)
where;
number of moles of sea water
number of moles of pure water
equating above equation 1 and 2; we have :
![\frac{n_2}{n_i}](https://tex.z-dn.net/?f=%5Cfrac%7Bn_2%7D%7Bn_i%7D)
![= \frac{0.0282}{0.9178}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B0.0282%7D%7B0.9178%7D)
![= 0.02901](https://tex.z-dn.net/?f=%3D%200.02901)
NOW, Molarity = ![\frac{moles of sea water}{mass of pure water }*1000](https://tex.z-dn.net/?f=%5Cfrac%7Bmoles%20of%20sea%20water%7D%7Bmass%20of%20pure%20water%20%7D%2A1000)
![= \frac{0.02901}{18}*1000](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B0.02901%7D%7B18%7D%2A1000)
![= 0.001616*1000](https://tex.z-dn.net/?f=%3D%200.001616%2A1000)
![= 1.616 M](https://tex.z-dn.net/?f=%3D%201.616%20M)
As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have ![\frac{1.616}{2} =0.808 M](https://tex.z-dn.net/?f=%5Cfrac%7B1.616%7D%7B2%7D%20%3D0.808%20M)
Answer:
![\large \boxed{\text{b. 1.5 h}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7Bb.%201.5%20h%7D%7D)
Explanation:
1. Calculate the rate constant
The integrated rate law for first order decay is
![\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt](https://tex.z-dn.net/?f=%5Cln%20%5Cleft%20%28%5Cdfrac%7BA_%7B0%7D%7D%7BA_%7Bt%7D%7D%5Cright%20%29%20%3D%20kt)
where
A₀ and A_t are the amounts at t = 0 and t
k is the rate constant
![\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Cln%20%5Cleft%20%28%5Cdfrac%7B500%7D%7B31%7D%5Cright%29%20%26%20%3D%20%26%20k%20%5Ctimes%206%5C%5C%5C%5C%5Cln%2016.1%20%26%20%3D%20%26%206k%5C%5C2.78%26%20%3D%26%206k%5C%5Ck%20%26%20%3D%20%26%20%5Cdfrac%7B2.78%7D%7B6%7D%5C%5C%5C%5C%26%20%3D%20%26%200.463%20%5Ctext%7B%20h%7D%5E%7B-1%7D%5C%5C%5Cend%7Barray%7D)
2. Calculate the half-life
![t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463 h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}](https://tex.z-dn.net/?f=t_%7B%5Cfrac%7B1%7D%7B2%7D%7D%20%3D%20%5Cdfrac%7B%5Cln2%7D%7Bk%7D%20%3D%20%5Cdfrac%7B%5Cln2%7D%7B%5Ctext%7B0.463%20%20h%7D%5E%7B-1%7D%7D%20%3D%20%5Ctextbf%7B1.5%20h%7D%5C%5C%5C%5C%20%5Ctext%7BThe%20half-life%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B1.5%20h%7D%7D%24%7D)
Answer:
Carbohydrates are energy-providing feed components composed of carbon, hydrogen, and oxygen. They should make up about 75 percent of an animal's diet. The energy they provide powers muscular movements. Carbohydrates also produce the body heat that helps keep the animal warm.
this is a dilution question where a certain volume is taken from a more concentrated solution and diluted to make a solution with a lower concentration.
the dilution forumla is as follows
c1v1 = c2v2
where c1 is concentration and v1 is volume of the more concetrated solution
c2 is concentration and v2 is volume of the diluted solution
substituting the values in the equation
18.0 M x V = 0.126 M x 16.5 L
V = 0.1155 L
volume of 115.5 mL is taken from the 18.0 M solution and diluted upto 16.5 L solution.