Chemistry, asked by shashimukund88, 9 months ago

Find the conc. Of H+ after mixing 15 ml, 0.1 M H2SO4 & 15 ml 0.1M NaOH?

Answers

Answered by mohishkhan9996
6

Answer:

hey.. mate ur ans.. is here

H2SO4 + 2NaOH = 2H2O + Na2SO4

mole H2SO4 = (0.005L)(1.0M) = 0.005mols H2SO4

mole NaOH = (0.010L)(0.1M) = 0.001mols NaOH

moles of H2SO4 consumed = 0.001mol NaOH (1mol H2SO4 / 2mol NaOH) = 0.0005 moles

H2SO4 left = 0.005-0.0005 = 0.0045moles H2SO4

NaOH left = 0 moles

Na2SO4 produce = 0.005moles

The hydrolysis of SO42- will be negligible so the pH contribution from this species can be neglected.

H2SO4 is a strong acid so in solution:

H2SO4 ----> 2H+ + SO42-

moles H+ = (0.0045moles H2SO4)(2moles H+ /1 moles H2SO4) = 0.09moles

[H+] = 0.09 mole / 0.015L = 0.6M H+

the total volume of the solution was 5ml + 10ml = 15mL, volume expansion was neglected.

pH = - log [H+] = - log (0.6) = 0.22

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Answered by Anonymous
1

Answer:

For 0.1M NaOH], pOH=-log(0.1)=1, which means the pH=13, (remember pH+pOH=14), but 13 has two sig. figures, so you must use the convention in figure 16.2. 2, and the answers are pH=1.0 for 1MHCL, and pH=13.0 for 1M NaOH.5 जून

Explanation:

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