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laiz [17]
3 years ago
5

Calculate the change in ph when 3.00 ml of 0.100 m hcl(aq) is added to 100.0 ml of a buffer solution that is 0.100 m in nh3(aq)

and 0.100 m in nh4cl(aq). a list of ionization constants can be found here.
Chemistry
1 answer:
nikklg [1K]3 years ago
7 0

The change in pH is calculated by:

pOH = Protein kinase B + log [NH4+]/ [NH3] 

Protein kinase B of ammonia = 4.74 

initial potential of oxygen hydroxide= 4.74 + log 0.100/0.100 = 4.74 
pH = 14 - 4.74=9.26 

moles NH4+ = moles NH3 = 0.100 L x 0.100 M = 0.0100 
moles H+ added = 3.00 x 10^-3 L x 0.100 M=0.000300 

NH3 + H+ = NH4+ 
moles NH3 = 0.0100 - 0.000300=0.00970 
moles NH4+ = 0.0100 + 0.000300=0.0103 

pOH = 4.74 + log 0.0103/ 0.00970= 4.77 
oH = 14 - 4.77 = 9.23 

the change is  = 9.26 - 9.23 =0.03 

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8 0
3 years ago
At rest, a neuron has a lower concentration of sodium than the surrounding fluid. The neuron also has a higher concentration of
pshichka [43]

Answer:

a. Remaining at rest requires the use of ATP.

Explanation:

The resting membrane potential is maintained by the sodium-potassium pump. The sodium potassium pump does this by actively pumping sodium ions out of the cell and potassium ions inside the cell in a ratio of 3:2. This movement of ions by the sodium-potassium pump is against their concentration gradient. In a neuron at rest, there are more sodium ions outside the cell than there are inside the cell. Also, there are are more potassium ions inside the cell than there are outside the cell. However, there are ion channels through which these ions enter and leave the cell. Sodium ion channels allow sodium to enter the cell following its concentration gradient, whereas, potassium ion channels allow potassium to leave the cell following its concentration gradient. However, more potassium ions leave the cell than do sodium ions enter the cell because of the higher permeability of the cell to potassium ions.

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6 0
3 years ago
Tris {(hoch2)3cnh2} is one of the most common buffers used in biochemistry. a solution is prepared by adding enough tris and 12
shusha [124]

Given that,

The concentration of TRIS = 0.30 M

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<span>pOH=14-pH=14-6.221 = 7.779</span>

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