Question 7 of 32 ncorrect Map General Chemistr University Science Books presente

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Question 7 of 32 ncorrect Map General Chemistr University Science Books presented by Sapling Lea Donald McQuarrie. Peter A Rock .Ethan Gallogly Calculate the pH of the resulting solution if 31.0 mL of 0.310 M HCI(aq) is added to (a) 41.0 mL of 0.310 M NaoH(aq). Number pH 11.491 (b) 21.0 mL of 0.410 M NaOH(aq). Number pH 1.72 There is additional feedback available! View this feedback by clicking on the bottom divider bar, Click on the divider bar again to hide the additional feedback. Close Previous ® Give Up & View olution Try Again Next Exit Explanation

Explanation / Answer

The reaction is NaOH+ HCl---> NaCl+ H2O

1mole requires 1 mole of HCL

moles of HCl= (31/1000)*0.31 =0.00961 moles

Moles of NaOH= 41*0.31/1000= 0.01271

out of this 0.00961 moles get neutralised, balance =0.01271-0.00961= 0.0031 moles

concentration = 0.0031*1000 / (31+41)= 0.0431 M

pOH- = -log [OH-] = -log [ 0.0431)= 1.4

pH+ pOH- =14

pH= 14-1.4= 12.6

In the second case

NaOH= 0.00861 mole

HCl = .00961 moles

HCl is excess by = 0.00961- 0.00861 =0.001 moles

concentration =0.001*1000/52=0.019M

pH =-log [ 0.019] =1.72

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