Calculate the pH of a Sr(OH)2 solution prepared by dissolving 0.3211 g of Sr(OH)2 and making up to 250.00 mL solution
Moles of Sr(OH)2 in 250ml.
"\\frac{mass}{molar\\space mass}=\\frac{0.3211}{121.63}=0.0026399736"
We can use the following relation to find molarity.
0.0026399736 moles "\\equiv" 250 ml
For 1000 ml we have
"\\frac{1000}{250}\u00d70.0026399736=0.01056 \\space moles"
pOH=-log [Sr (OH)2]
pH for bases is essentially 14-pOH.
"pH=14-(-log\\space0.01056)=12.024"
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