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In: Chemistry

In the titration of 0.010 L of 2.0 M NaOH with 1.0 M HI, what is...

In the titration of 0.010 L of 2.0 M NaOH with 1.0 M HI, what is the pH when 0.015 L of HI have been added?

Expert Solution

HI NaOH

MA = 1M MB = 2M

VA = 0.015L VB = 0.010M

M = MBVB-MAVA/VA+VB MBVB>MAVA

= 2*0.010 -1*0.015/0.01+0.015

= 0.005/0.01+0.015

= 0.005/0.025 = 0.2 M

[OH^-] = M = 0.2M

POH = -log[OH^-]

= -log0.2

= 0.6989

PH = 14-POH

= 14-0.6989

= 13.3011 >>>>answer

= -log0.2

queen_honey_blossom answered 2 years ago

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