Question

Calculate the ph of a solution formed by mixing 100 ml of 0.1m hcl and 9.9ml of 1m naoh

Answer 1

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Answer 2

pH of a solution formed by mixing 100 ml of 0.1m HCl and 9.9ml of 1m NaOH is 3.0

Explanation:

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

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

Molarity of HCl solution = 0.1 M

Volume of HCl solution = 100 ml = 0.1 L

$\text{Moles of HCl}={0.1M\times 0.1L=0.01mol$    

Putting values in equation 1, we get:

Molarity of NaOH solution = 1 M

Volume of NaOH solution = 9.9 ml = 0.0099 L

$\text{Moles of NaOH}={1M\times 0.0099L=0.0099mol$   The chemical equation for the reaction of NaOH with HCl follows:

$HCl+NaOH\rightarrow NaCl+H_2O$

Moles of $H^+$ ion =0.01 moles

Moles of $OH^-$ ion = 0.0099 moles

1 mole of $OH^-$ ion will react with 1 mole of $H^+$ ion

0.0099 moles of $OH^-$ ion will react with=$\frac{1}{1}\times 0.0099=0.0099$ moles of $H^+$ ion

Moles of $H^+$ left = (0.01-0.0099) = 0.0001

Molarity of $[H^+]=\frac{\text {moles left}}{\text {total volume}}=\frac{0.0001}{0.1099L}=0.0009M$

$pH=-log[H^+]=-log[0.0009]=3.0$

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