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How Many Moles Of HCl Need To Be Added To 150.0 ML Of 0.50 M NaZ To Have A Solution With A PH Of 6.50

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Eduard22sly

The number of mole of HCl needed for the solution is 1.035×10¯³ mole

How to determine the pKa

We'll begin by calculating the pKa of the solution. This can be obtained as follow:

  • Equilibrium constant (Ka) = 2.3×10¯⁵
  • pKa =?

pKa = –Log Ka

pKa = –Log 2.3×10¯⁵

pKa = 4.64

How to determine the molarity of HCl

  • pKa = 4.64
  • pH = 6.5
  • Molarity of salt [NaZ] = 0.5 M
  • Molarity of HCl [HCl] =?

pH = pKa + Log[salt]/[acid]

6.5 = 4.64 + Log[0.5]/[HCl]

Collect like terms

6.5 – 4.64 = Log[0.5]/[HCl]

1.86 = Log[0.5]/[HCl]

Take the anti-log

0.5 / [HCl] = anti-log 1.86

0.5 / [HCl] = 72.44

Cross multiply

0.5 = [HCl] × 72.44

Divide both side by 72.44

[HCl] = 0.5 / 72.4

[HCl] = 0.0069 M

How to determine the mole of HCl

  • Molarity of HCl = 0.0069 M
  • Volume = 150 mL = 150 / 1000 = 0.15 L
  • Mole of HCl =?

Mole = Molarity x Volume

Mole of HCl = 0.0069 × 0.15

Mole of HCl = 1.035×10¯³ mole

Complete question

How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl

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