calculate the ph of 5×10^-4 m ca(oh)2
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Explanation:
Answer
Molarity =10
−8
∴ Normality =2×10
−8
pOH=−log[
−
OH]=−log[2×10
−8
]=−log2+[−log10
−8
]=−0.3010+8=7.69
∵pH+pOH=14
∴pH=14−pOH=14−7.69=6.31
∴ pH of 10
−8
molar Ca(OH)
2
=6.31
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