Answer:
0.12M
Explanation:
A balanced equation for the reaction will go a great deal in obtaining our desired result. So, let us write a balanced equation for the reaction
HCl + NaOH —> NaCl + H2O
From the above equation,
nA (mole of the acid) = 1
nB (mole of the base) = 1
Data obtained from the question include:
Vb (volume of the base) = 30mL
Mb (Molarity of the base) = 0.1M
Va (volume of the acid) = 25mL
Ma (Molarity of the acid) =?
The molarity of the acid can be obtained as follow:
MaVa/MbVb = nA/nB
Ma x 25/ 0.1 x 30 = 1
Cross multiply to express in linear form
Ma x 25 = 0.1 x 30
Divide both side by 25
Ma = (0.1 x 30) / 25
Ma = 0.12M
The molarity of the acid is 0.12M
Answer:
2,760 grams NaCl
Explanation:
To find grams of NaCl, you need to (1) convert moles of Na to moles of NaCl (via mole-to-mole ratio from reaction) and (2) convert moles of NaCl to grams (via molar mass from periodic table). The final answer should have 3 significant figures based on the given measurement.
2 Na + Cl₂ --> 2 NaCl
Molar Mass (NaCl) = 22.99 g/mol + 35.45 g/mol
Molar Mass (NaCl) = 58.44 g/mol
47.2 moles Na 2 moles NaCl 58.44 grams
---------------------- x --------------------------- x ------------------------- =
2 moles Na 1 mole NaCl
= 2,758.368 grams NaCl
= 2,760 grams NaCl
Answer:
4.22
Explanation:
pH stands for potential hydrogen. The letter “p” denotes potential and the letter “H” denotes hydrogen.
pH helps to find the acidity or alkalinity of an aqueous solution.
The number of hydrogen ions (protons) present in a solution is determined by the pH scale.
A pH greater than 7 makes the water more alkaline and a pH less than 7 makes the water more acidic.
![pH=-\log [H^+]=-\log [0.00006]=4.22](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D%3D-%5Clog%20%5B0.00006%5D%3D4.22)
This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.
Answer:
0.22 mol HClO, 0.11mol HBr.
0.25mol NH₄Cl, 0.12 mol HCl
Explanation:
A buffer is defined as a mixture in solution between weak acid and its conjugate base or vice versa.
Potassium hypochlorite (KClO) could be seen as conjugate base of HClO (Weak acid). That means the addition of <em>0.22 mol HClO </em>will convert the solution in a buffer. HBr reacts with KClO producing HClO, thus, <em>0.11mol HBr</em> will, also, convert the solution in a buffer. 0.23 mol HBr will react completely with KClO and in the solution you will have only HClO, no a buffering system.
Ammonia (NH₃) is a weak base and its conjugate base is NH₄⁺. That means the addition of <em>0.25mol NH₄Cl</em> will convert the solution in a buffer. Also, NH₃ reacts with HCl producing NH₄⁺. Thus, addition of<em> 0.12 mol HCl</em> will produce NH₄⁺. 0.25mol HCl consume all NH₃.
