Question

In: Chemistry

A 340.0 −mL buffer solution is 0.140 M in HF and 0.140 M in NaF. What...

A 340.0 −mL buffer solution is 0.140 M in HF and 0.140 M in NaF. What mass of NaOH could this buffer neutralize before the pH rises above 4.00? And, If the same volume of the buffer was 0.370 M in HF and 0.370 M in NaF, what mass of NaOHcould be handled before the pH rises above 4.00? Please show all steps. Thank you

Solutions

Expert Solution

Sol 1.

Let the moles of NaOH added be x  

initial moles of HF = 0.140 × 340.0 / 1000 = 0.0476 mol

initial moles of NaF = 0.140 × 340.0 / 1000 = 0.0476 mol

After addition of moles of NaOH , moles of NaF increases and moles of HF decreases

Final moles of HF = 0.0476 - x

Final moles of NaF = 0.0476 + x

Also , pKa of HF = 3.2   

So , Using Henderson - Hasselbalch equation ,

pH = pKa + log ( Final moles of NaF / Final moles of HF )

4 = 3.2 + log ( (0.0476 + x ) / ( 0.0476 - x ) )

log ( (0.0476 + x ) / ( 0.0476 - x ) ) = 4 - 3.2 = 0.8

(0.0476 + x ) / ( 0.0476 - x ) = 100.8 = 6.3096

0.0476 + x = 0.3003 - 6.3096x

x = 0.2527 / 7.3096 = 0.0346 mol

As molar mass of NaOH = 40 g / mol

So , mass of NaOH = 0.0346 × 40 =   1.384 g

Sol 2.

Let the moles of NaOH added be x  

initial moles of HF = 0.370 × 340.0 / 1000 = 0.1258 mol

initial moles of NaF = 0.370 × 340.0 / 1000 = 0.1258 mol

After addition of moles of NaOH , moles of NaF increases and moles of HF decreases

Final moles of HF = 0.1258 - x

Final moles of NaF = 0.1258 + x

Also , pKa of HF = 3.2   

So , Using Henderson - Hasselbalch equation ,

pH = pKa + log ( Final moles of NaF / Final moles of HF )

4 = 3.2 + log ( (0.1258 + x ) / ( 0.1258 - x ) )

log ( (0.1258 + x ) / ( 0.1258 - x ) ) = 4 - 3.2 = 0.8

(0.1258 + x ) / ( 0.1258 - x ) = 100.8 = 6.3096

0.1258 + x = 0.7937 - 6.3096x

x = 0.6679 / 7.3096 = 0.0914 mol

As molar mass of NaOH = 40 g / mol

So , mass of NaOH = 0.0914 × 40 =   3.656 g


Related Solutions

A 340.0 −mL buffer solution is 0.170 M in HF and 0.170 M in NaF. Part...
A 340.0 −mL buffer solution is 0.170 M in HF and 0.170 M in NaF. Part A What mass of NaOH could this buffer neutralize before the pH rises above 4.00? Express your answer using two significant figures. m =   g   Part B If the same volume of the buffer was 0.350 M in HF and 0.350 M in NaF, what mass of NaOH could be handled before the pH rises above 4.00? Express your answer using two significant figures.
#16.51 A 340.0 −mL buffer solution is 0.130 M in HF and 0.130 M in NaF....
#16.51 A 340.0 −mL buffer solution is 0.130 M in HF and 0.130 M in NaF. What mass of NaOH could this buffer neutralize before the pH rises above 4.00? ANSWER: 0.98 If the same volume of the buffer was 0.340 M in HF and 0.340 M in NaF, what mass of NaOHcould be handled before the pH rises above 4.00?
A 340.0 mL buffer solution is 0.150M in HF and 0.150M in NaF. Part A What...
A 340.0 mL buffer solution is 0.150M in HF and 0.150M in NaF. Part A What mass of NaOH could this buffer neutralize before the pH rises above 4.00? Part B If the same volume of the buffer were 0.330 M in HF and 0.330 M in NaF, what mass of NaOH could be handled before the pH rises above 4.00? Express answer using two significant figures.
A 360.0 −mL buffer solution is 0.160 M in HF and 0.160 M in NaF. What...
A 360.0 −mL buffer solution is 0.160 M in HF and 0.160 M in NaF. What mass of NaOH could this buffer neutralize before the pH rises above 4.00? If the same volume of the buffer was 0.360 M in HF and 0.360 M in NaF, what mass of NaOH could be handled before the pH rises above 4.00?
A 360.0 −mL buffer solution is 0.170 M in HF and 0.170 M in NaF. A)...
A 360.0 −mL buffer solution is 0.170 M in HF and 0.170 M in NaF. A) What mass of NaOH could this buffer neutralize before the pH rises above 4.00? B) If the same volume of the buffer was 0.360 M in HF and 0.360 M in NaF, what mass of NaOH could be handled before the pH rises above 4.00? I keep getting 1.8 g and 3.5 g, which is wrong.
A 360.0 −mL buffer solution is 0.130 M in HF and 0.130 M in NaF. Part...
A 360.0 −mL buffer solution is 0.130 M in HF and 0.130 M in NaF. Part A) What mass of NaOH can this buffer neutralize before the pH rises above 4.00? Part B) If the same volume of the buffer was 0.360 M in HF and 0.360 M in NaF, what mass of NaOH could be handled before the pH rises above 4.00?
A 25.0 mL buffer solution is 0.350 M in HF and 0.150 M in NaF. Calculate...
A 25.0 mL buffer solution is 0.350 M in HF and 0.150 M in NaF. Calculate the pH of the solution after the addition of 37.5 mL of .200 M HCl. The pKa for HF is 3.46.
A 350 mL buffer solution containing 0.15 M HF and 0.150 M NaF is reacted with...
A 350 mL buffer solution containing 0.15 M HF and 0.150 M NaF is reacted with sodium hydroxide. What mass of NaOH can this buffer neutralize before the pH rises above 4.0? If the same volume of buffer was 0.35 M in HF and 0.35 M in NaF then what mass of NaOH could be handled before the pH rises above 4.0?
A 350.0 mL buffer solution is 0.170M in HF and 0.170M in NaF. A. What mass...
A 350.0 mL buffer solution is 0.170M in HF and 0.170M in NaF. A. What mass of NaOH could this buffer neutralize before the pH rises above 4.00? B. If the same volume of the buffer was 0.370M in HF and 0.370M in NaF, what mass of NaOH could be handled before the pH rises above 4.00?
A buffer solution that is 0.403 M in HF and 0.403 M in NaF has a...
A buffer solution that is 0.403 M in HF and 0.403 M in NaF has a pH of 3.14. (1) The addition of 0.01 mol of H3O+ to 1.0 L of this buffer would cause the pH to _________increase slightly increase by 2 units decrease slightly decrease by 2 units not change. (2) The capacity of this buffer for added H3O+could be increased by the addition of 0.148 mol_________of the weak acid of the salt. Consider these compounds: A. CaCrO4...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT