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Finger [1]
3 years ago
7

Tomato juice has a pH of 4.20. Calculate the [H3O+] and [OH–] in tomato juice.

Chemistry
1 answer:
Free_Kalibri [48]3 years ago
8 0
<span>[H3O+] = 10^(-pH) = 10^(-4.20) = 6.3 x 10^-5 M

pOH = 14 - pH = 14 - 4.20 = 9.80

[OH-] = 10^(-pOH) = 10^(-9.80) = 1.6 x 10^-10 M
</span>
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You decided to prepare a phosphate buffer from solid sodium dihydrogen phosphate (NaH2PO4) and disodium hydrogen phosphate (Na2H
KIM [24]

Answer:

For disodium hydrogen phosphate:

5.32g Na2HPO4

For sodium dihydrogen phosphate:

7.65g Na2HPO4

Explanation:

First, you have to put all the data from the problem that you going to use:

-NaH2PO4 (weak acid)

-Na2HPO4 (a weak base)

-Volume = 1L

-Buffer pH = 7.00

-Concentration of [NaH2PO4 + Na2HPO4] = 0.100 M

What we need to find the pKa of the weak acid, in this case NaH2PO4, for that you need to find the Ka (acid constant) of NaH2PO4, and for this we use the pKa of the phosphoric acid as follow:

H3PO4 = H2PO4 + H+    pKa1 = 2.14

H2PO4 = HPO4 + H+       pKa2 = 6.86

HPO4 = PO4 + H+      pKa3 = 12.4

So, for the preparation of buffer, you need to use the pKa that is near to the value of the pH that you want, so the choice will be:

pKa2= 6.86

Now we going to use the Henderson Hasselbalch equation for the pH of a buffer solution:

pH = pKa2 + log [(NaH2PO4)/(Na2HPO4)]

The solution of the problem is attached to this answer.

Download odt
7 0
3 years ago
A 32.2 g iron rod, initially at 21.9 C, is submerged into an unknown mass of water at 63.5 C. in an insulated container. The fin
lorasvet [3.4K]

Answer : The mass of the water in two significant figures is, 3.0\times 10^1g

Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron metal = 0.45J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of iron metal = 32.3 g

m_2 = mass of water = ?

T_f = final temperature of mixture = 59.2^oC

T_1 = initial temperature of iron metal = 21.9^oC

T_2 = initial temperature of water = 63.5^oC

Now put all the given values in the above formula, we get

32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC

m_2=30.16g\approx 3.0\times 10^1g

Therefore, the mass of the water in two significant figures is, 3.0\times 10^1g

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