please solve the above question fast
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the ph is 12.65 it so easy
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sonu806:
but say hoe
I mean how
Plz say
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Using formula
M=(m2v2-m1v1)/v1+v2 (here v1=v2)
M=(0.1-0.01)v/2v
M=0.045
here as NaOH is in execess ... -log M will give pOH
pOH=-log 0.045
pOH=1.64 (highly aproxx)
pH=14-pOH
pH=12.65 (Highly aproxx)
hope this helps
M=(m2v2-m1v1)/v1+v2 (here v1=v2)
M=(0.1-0.01)v/2v
M=0.045
here as NaOH is in execess ... -log M will give pOH
pOH=-log 0.045
pOH=1.64 (highly aproxx)
pH=14-pOH
pH=12.65 (Highly aproxx)
hope this helps
but y negative
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