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Grade: 11
        Question no 78  please answer as soon as possible thanks
3 months ago

Answers : (1)

Arun
8936 Points
							
According to the given question, HCl + NaOH → NaCl + H₂

As per the details that are given in the question, we can calculate the following:-

No. of moles of HCl = (40 ml) (0.1 mol/L)

= (0.04 L) (0.1 mol/L) = 0.004 moles

No. of moles of NaOH = (10 ml) (0.45 mol/L)

= (0.01 L) (0.45 mol/L) = 0.0045 moles

Therefore, excess amount of NaOH will be = 0.0045 moles - 0.004 moles

= 0.0005 moles 

Thus, the total volume will be = 0.04 L + 0.01 L = 0.05 L 

Molarity of OH will be = 0.0005 moles/0.05 L = 0.01 M

pOH = -log [OH] 

= -log 0.01 

Therefore, pOH = 2 

so, pH = 14 - pOH 

Thus, pH of the above solution is = 12
2 months ago
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