Question

100ml of 0.05M h2so4 + 500ml of 0.1M hcl +400ml of 0.1M ca(oh)2 .fin the ph.explain please

Answer 1

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Answer 2

The pH of the final solution is 12.3

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Moles}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}$     .....(1)

a) $\text{Moles of} H_2SO_4={0.05}\times{0.1}=0.005moles$

1 mole of $H_2SO_4$ gives 2 moles of $H^+$

Thus 0.005 moles of $H_2SO_4$ give =$\frac{2}{1}\times 0.005=0.01$ moles of $H^+$

b) $\text{Moles of} HCl={0.1}\times{0.5}=0.05moles$

1 mole of $HCl$ gives 1 mole of $H^+$

Thus 0.05 moles of $HCl$ give =$\frac{1}{1}\times 0.05=0.05$ moles of $H^+$

c) $\text{Moles of} Ca(OH)2={0.1}\times{0.4}=0.04moles$

1 mole of $Ca(OH)_2$ gives 2 moles of $OH^-$

Thus 0.04 moles of $Ca(OH)_2$ give =$\frac{2}{1}\times 0.04=0.08$ moles of $OH^-$

Total Moles of $H^+$ = 0.01 + 0.05 = 0.06

Total Moles of $OH^-$ = 0.08

$H^++OH^-\rightarrow H_2O$  

According to stoichiometry :

1 moles of $H^+$ require 1 mole of $OH^_$

Thus 0.06 moles of $H^+$ will require=$\frac{1}{1}\times 0.06=0.06moles$  of $OH^-$

Moles of $OH^-$ left = 0.08-0.06 = 0.02moles

Total volume = 100ml + 500ml + 400 ml =1000 ml = 1L

$[OH^-]=\frac{moles}{\text {Total Volume in L}}=\frac{0.02}{1L}=0.02M$

$pOH=-log[OH^-]=-log[0.02]=1.7$

$pH+pOH=14$

$pH=14-1.7=12.3$

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