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4 years ago

How many moles of potassium hydroxide (KOH) are needed to completely neutralize 1.56 mol phosphoric acid (H3PO4)?

Chemistry

2 answers:

photoshop1234 [79]4 years ago

4 0

.56 mol H3PO4 x (3 mol KOH/1 mol H3PO4)=4.68 mol KOH.

bekas [8.4K]4 years ago

4 0

Answer : The number of moles of potassium hydroxide (KOH) are needed to completely neutralize are 4.68 moles.

Explanation :

The balanced chemical reaction will be:

$3KOH+H_3PO_4\rightarrow K_3PO_4+3H_2O$

From the balanced chemical reaction we conclude that,

As, 1 mole of $H_3PO_4$ neutralizes 3 moles of KOH

So, 1.56 mole of $H_3PO_4$ neutralizes $1.56\times 3=4.68$ moles of KOH

Thus, the number of moles of potassium hydroxide (KOH) are needed to completely neutralize are 4.68 moles.

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julia-pushkina [17]

Answer: The $K_{sp}$ for calcium hydroxide is $5.324\times 10^{-6}$

Explanation:

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$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

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The concentration of $Ca(OH)_2$ comes out to be 0.011 M.

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$Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-$

The expression for solubility constant for this reaction follows:

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Putting the values in above equation, we get:

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