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3 years ago

Helppppp will mark brainlest

Chemistry

1 answer:

ss7ja [257]3 years ago

5 0

Answer:

if you will decrease the HCO3- so the less H+ ion will be form and reaction will more likely shift to product and when HCO3- decreases pH value increases and vica versa for the increasing HCO3 the more H+ ion will be form and reaction will shift to product and the pH value will decreases!!

I've doubt in reaction shift coz whatever is the amount of HCO3- this is completely gonna form number of H+ so reaction shift may or may not be same!!

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Complete this equation for the dissociation of the electrolyte NaCl(aq). Omit water from the equation because it is understood t

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1. Answer;

Equation;

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-The polar property of water allows ionic compounds such as sodium chloride to dissolve in water. (sodium chloride is an ionic compound).

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2. Answer;

Equation;

Fe(ClO4)3(aq) → Fe 3+(aq) + 3ClO4 - (aq)

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Iron (iii) perchlorate (Fe(ClO4)3 dissociates in water into iron (iii) ion and perchlorate ion. Iron (iii) ion is positively charged while the perchlorate ion is negatively charged, they are pulled apart by the polar water molecules, a process called dissociation.

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3 years ago

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anastassius [24]

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As a technician in a large pharmaceutical research firm, you need to produce 250. mL of a potassium dihydrogen phosphate buffer

Reika [66]

Answer:

We will need 147.772 mL of KH2PO4 to make this solution

Explanation:

For this case we can give the following equation:

H2PO4 - ⇄ H+ + HPO42-

With following pH- equation:

pH = pKa + log [HPO42-]/[H2PO4-]

7.05 = 7.21 + log [HPO42-]/[H2PO4-]

-0.16 = log [HPO42-]/[H2PO4-]

10^-0.16 = [HPO42-]/[H2PO4-]

0.6918 = [HPO42-]/[H2PO4-]

Let's say the volume of HPO42-= x then the volume of H2PO4- will be 250 mL - x

Since both have a concentration of 1M = 1 mol /L

If we plug this in the equation 0.6918 = [HPO42-]/[H2PO4-]

0.6918 = x / (250 - x)

0.6918*250 - 0.6918x = x

172.95 = 1.6918x

x = 102.228 mL

The volume of HPO42- = 102.228 mL

Then the volume of H2PO4- = 250 - 102.228 = 147.772 mL

To control this we can plug this in the pH equation

7.05 = 7.21 + log [HPO42-]/[H2PO4-]

7.05= 7.21 + log (102.228 / 147.772) = 7.05

We will need 147.772 mL of KH2PO4 to make this solution

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