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Law Incorporation [45]
3 years ago
7

What is the ph of a buffer prepared by adding 1.00 l of 1.0 m hcℓ to 750 ml of 1.5 m nahcoo? ka = 1.7 x 10-4?

Chemistry
1 answer:
alexandr1967 [171]3 years ago
5 0
Ph = 2.837 ( solution is attached ) 

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Explanation:

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What are the different characteristics that can be used to identify an unknown mineral
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The freezing temperature of water is zero on both the "K" and "C" scales. True False
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Which model of election configuration is the most accurate? The quantum mechanical atomic model The Rutherford atomic model The
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Quantum mechanical atomic model.

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4 years ago
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calculate the molarity of MgCl2 in the following solution: 5.34 g of MgCl2 is dissolved and diluted to 214 mL of solution.
anyanavicka [17]
<h3><u>Ⲁⲛ⳽ⲱⲉⲅ</u><u>:</u></h3>

\quad\hookrightarrow\quad \sf {0.262M }

<h3><u>Ⲋⲟⳑⳙⲧⳕⲟⲛ :</u></h3>

Molarity is used to measure the concentration of a solution , so it is also as molar concentration. It is denoted as M or Mol/L

<u>We </u><u>are </u><u>given </u><u>that </u><u>:</u>

  • Weight of \sf MgCl_{2} = 5.34g
  • Volume of solution = 214 ml , or 0.214 L

The molar mass of magnesium chloride ( \sf MgCl_{2} ) is 95.21 g / mol

We can calculate the molarity of the solution by dividing the number of moles of solute by volume of solvent in liter ,i.e:

\quad\longrightarrow\quad \sf  {M = \dfrac{n}{V} } ‎ㅤ‎ㅤ‎ㅤ⸻( 1 )

<em>Where,</em><em> </em>

  • M = molarity
  • n = number of moles
  • V = Volume

We can calculate the number of moles by dividing the actual mass by its molar mass ,i.e:

\quad\longrightarrow\quad \sf { n = \dfrac{w}{m}}‎ㅤ‎ㅤ‎ㅤ‎⸻ ( 2 )

<em>W</em><em>here,</em>

  • n = number of moles
  • m = molar mass
  • w = actual mass

<u>Therefore</u><u>,</u>

\implies\quad \tt {n =\dfrac{w}{m} }

\implies\quad \tt { n =\dfrac{5.35\: g}{95.21\: g /mol}}

\implies\quad{\pmb{ \tt {n = 0.056 mol}} }

<u>P</u><u>utting </u><u>the </u><u>values </u><u>in </u><u>equation </u><u>(</u><u> </u><u>1</u><u> </u><u>)</u><u>:</u>

\implies\quad \tt {M=\dfrac{n}{V} }

\implies\quad \tt { M =\dfrac{0.056\:mol}{0.214\:L}}

\implies\quad\underline{\pmb{ \tt { M = 0.262 \:M }}}

7 0
2 years ago
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