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vagabundo [1.1K]

3 years ago

13

If 0.580 moles of a monoprotic weak acid (ka = 7.4 10-5 is titrated with naoh, what is the ph of the solution at the half-equiva

lence point?

1 answer:

Yuri [45]3 years ago

7 0

<span><span>S is for soil,</span><span>cl (sometimes c) represents climate,</span><span>o organisms including humans,</span><span>r relief,</span><span>p parent material, or lithology, and</span><span>t time.</span></span>

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You are given a solution of HCOOH (formic acid) with an approximate concentration of 0.20 M and you will titrate this with a 0.1

jeka57 [31]

Answer:

$\boxed{\text{36 mL}}$

Explanation:

1. Write the balanced chemical equation.

$\rm HCOOH + NaOH $ \longrightarrow$ HCOONa + H$_{2}$O$

2. Calculate the moles of HCOOH

$\text{Moles of HCOOH} =\text{20.00 mL HCOOH } \times \dfrac{\text{0.20 mmol HCOOHl}}{\text{1 mL HCOOH}} = \text{4.00 mmol HCOOH}$

3. Calculate the moles of NaOH.

$\text{Moles of NaOH = 4.00 mmol HCOOH } \times \dfrac{\text{1 mmol NaOH} }{\text{1 mmol HCOOH}} = \text{4.00 mmol NaOH}$

4. Calculate the volume of NaOH

$c = \text{4.00 mmol NaOH } \times \dfrac{\text{1 mL NaOH }}{\text{0.1105 mmol NaOH }} = \textbf{36 mL NaOH }\\\\\text{The titration will require }\boxed{\textbf{36 mL of NaOH}}$

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