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N76 [4]
3 years ago
8

Fill in the coefficients that will balance the following reaction: a0NH4Cl + a1Ag3PO4 → a2AgCl + a3(NH4)3PO4

Chemistry
1 answer:
rosijanka [135]3 years ago
8 0
The equation of the reaction before balancing is 
a0NH₄Cl + a1Ag₃PO₄ → a2AgCl + a3(NH₄)₃PO₄
PO₄³⁻ ion is balanced. 
on the left side, theres 1 (NH₄⁺) ion and right side 3 (NH₄⁺) ions. Therefore if we put the coefficient for NH₄Cl, we will obtain the following equation 
3 NH₄Cl + a1Ag₃PO₄ → a2AgCl + a3(NH₄)₃PO₄
3 Ag⁺ ions on the left side and 1 Ag⁺ ion on the right side, so if we put the coefficient of AgCl as 3, following equation obtained 
3 NH₄Cl + a1Ag₃PO₄ → 3 AgCl + a3(NH₄)₃PO₄
Cl⁻ ions are also balanced now, 3 on either side.
a1 and a3 are 1 as those compounds are as it is, so coefficient is 1 for both 


balanced equation is as follows 
3 NH₄Cl + Ag₃PO₄ → 3 AgCl + (NH₄)₃PO₄
coefficients are 
a0 - 3
a1 - 1
a2 - 3
a3 - 1
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Calculate either [ H 3 O + ] or [ OH − ] for each of the solutions.
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Answer: Solution A : [H_3O^+]=0.300\times 10^{-7}M

Solution B : [OH^-]=0.107\times 10^{-5}M

Solution C : [OH^-]=0.177\times 10^{-10}M

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

pH=-\log[H_3O^+]

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pH+pOH=14

[H_3O^+][OH^-]=10^{-14}

a. Solution A: [OH^-]=3.33\times 10^{-7}M

[H_3O^+]=\frac{10^{-14}}{3.33\times 10^{-7}}=0.300\times 10^{-7}M

b. Solution B : [H_3O^+]=9.33\times 10^{-9}M

[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M

c. Solution C : [H_3O^+]=5.65\times 10^{-4}M

[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M

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