Answer:
The pH value of the mixture will be 7.00
Explanation:
Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,
![pH=pK_{a} + log(\frac{[Base]}{[Acid]})](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%20%2B%20log%28%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%29)
According to the given conditions, the equation will become as follow
![pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%20%2B%20log%28%5Cfrac%7B%5BNa_%7B2%7DHPO_%7B4%7D%20%5D%7D%7B%5BNaH_%7B2%7DPO_%7B4%7D%5D%7D%29)
The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.
Placing all the given data we obtain,
B) False- It has seven
A hexagon would have 6.
Answer:
a) Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)
b) 3.14g must be added
Explanation:
a) For the reaction:
Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) + H₂O(l) + NH₃(g)
As you see, there are 8 moles of water in reactants and 2 moles of oxygen in octahydrate, thus, water moles must be 10:
Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + NH₃(g)
To balance hydrogens, the other coefficients are:
Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)
b) As you see in the balanced reaction, 1 mole of barium hydroxide octahydrate reacts with 2 moles of NH₄SCN. 6.5g of Ba(OH)₂.8H₂O are:
6.5 g × (1mol / 315.48g) =<em> 0.0206moles of Ba(OH)₂.8H₂O</em>. Thus, moles of NH₄SCN that must be used for a complete reaction are:
0.0206moles of Ba(OH)₂.8H₂O × ( 2 mol NH₄SCN / 1 mol Ba(OH)₂.8H₂O) = <em>0.0412moles of NH₄SCN</em>. In grams:
0.0412moles of NH₄SCN × ( 76.12g / 1mol) = <em>3.14g must be added</em>
It is the boilimg point now aa
