Question

0.24g sample of mg reacts with 80.0 ml of hcl solution whose ph is -0.30. ph after all mg has reacted is​

Answer 1

Answer:

Mg(aq)+2HCl(aq)⟶MgCl

2

(aq)+H

2

Number of moles of Mg is

24.3

1.458

=0.06 moles =60 M moles

Molarity of HCl=3M

⇒3×80=240 M moles

HCl left after reaction= 240−(60×2)=120 M moles

Hence molarity of solution=

80

120

=1.5M

∴pH=−log[H

+

]=−log1.5=−0.176

∴ Change in pH=−0.176−(−0.477)=0.3

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