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Calculate the pH of a 365.8-mL sample of 0.007 M Ba(OH)2 solution. Assume complete dissociation. Provide...

Calculate the pH of a 365.8-mL sample of 0.007 M Ba(OH)₂ solution. Assume complete dissociation. Provide your answer to two decimal places and without units.

Expert Solution

Ba(OH)₂ ---> Ba⁺² + 2(OH)^-

0.007 M 0.007 M 0.014 M

Given,

Volume of sample = 365.8 ml

So the concentration of OH^- ions = 0.014 * 365.8 / 1000

= 5.1212 * 10^-3 M

Since,

pH = 14 – pOH ---(1)

pOH = -log( [OH^-] ) ---(2)

so putting the value in equation 2 and then in equation 1

pOH = -log( 5.1212 * 10^-3 )

pOH = 3 – log ( 5.1212 )

pOH = 3 – 0.7093

pOH = 2.2906

and now by equation 1

pH = 14 – pOH

pH = 14 – 2.2906

pH = 11.70 (this is the required answer )

queen_honey_blossom answered 2 years ago

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