How much pure water must mixed with 8 pints of 50 developer to produce a mixture that is 24%?

1 answer:

6 0

8:24 = 0.333(3) pints is one percent
0.333(3)* 100=33.333(3) pints will be 24% mixture with the water
33.3333-8=25.333 pints of water is required for producing 24% mixture

25.3333 pints of pure water and 8 pints of juce.

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In a titration of 47.41 mL of 0.3764 M ammonia with 0.3838 M aqueous nitric acid, what is the pH of the solution when 47.41 mL +

Volgvan

Answer: The pH of the solution is 1.136

Explanation:

To calculate the moles from molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

For ammonia:

Molarity of ammonia = 0.3764 M

Volume of ammonia = 47.41 mL = 0.04741 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.3764mol/L=\frac{\text{Moles of ammonia}}{0.04741L}\\\\\text{Moles of ammonia}=0.01784mol$

For nitric acid:

Molarity of nitric acid = 0.3838 M

Volume of ammonia = (47.41 + 10.00) mL = 57.41 mL= 0.05741 L

Putting values in above equation, we get:

$0.3838mol/L=\frac{\text{Moles of nitric acid}}{0.05741L}\\\\\text{Moles of nitric acid}=0.02203mol$

After the completion of reaction, amount of nitric acid remained = 0.022 - 0.0178 = 0.0042 mol

For the reaction of ammonia with nitric acid, the equation follows:

$NH_3+HNO_3\rightarrow NH_4NO_3$

At $t=0$ 0.0178 0.022

Completion 0 0.0042 0.0178

As, the solution of the reaction is made from strong acid which is nitric acid and the conjugate acid of weak base which is ammonia. So, the pH of the reaction will be based totally on the concentration of nitric acid.

To calculate the pH of the reaction, we use the equation:

$pH=-\log[H^+]$